💡Okay, so I have this problem about a parabola and a line intersecting it. Let me try to understand what's being asked here. The parabola is given by the equation ( y^2 = 2px ) where ( p > 0 ). I remember that for a parabola in this form, the focus is at ( ( frac{p}{2}, 0 ) ). So, the focus ( F ) is located on the x-axis, halfway between the vertex at the origin and the point ( (p, 0) ).Now, there's a line ( l ) that passes through this focus ( F ) and makes a slope angle of ( 60^circ ). I think the slope angle is the angle the line makes with the positive x-axis, measured counterclockwise. So, a ( 60^circ ) angle would mean the slope of the line is ( tan(60^circ) ), which is ( sqrt{3} ). Therefore, the equation of line ( l ) should be ( y = sqrt{3}(x - frac{p}{2}) ), since it passes through ( F ).This line intersects the parabola ( C ) at two points ( A ) and ( B ). The problem mentions that ( A ) is in the first quadrant and ( B ) is in the fourth quadrant. That makes sense because the parabola opens to the right, so points in the first and fourth quadrants would be on either side of the x-axis.Additionally, the line ( l ) intersects the axis of the parabola at point ( P ). The axis of the parabola ( y^2 = 2px ) is the x-axis. So, to find point ( P ), I need to find where line ( l ) intersects the x-axis. That occurs when ( y = 0 ). Plugging that into the equation of line ( l ), we get ( 0 = sqrt{3}(x - frac{p}{2}) ), which simplifies to ( x = frac{p}{2} ). Wait, that's the same as the focus ( F ). Hmm, that seems odd. Maybe I made a mistake here.Wait, no, actually, the axis of the parabola is the x-axis, so any line intersecting the axis would be at ( y = 0 ). But since the line ( l ) already passes through the focus ( F ), which is on the x-axis, point ( P ) is just ( F ). But the problem says the line intersects the axis at point ( P ). Maybe I need to double-check this.Wait, perhaps the axis of the parabola is considered to be the line of symmetry, which is the x-axis here. So, the line ( l ) intersects the x-axis at ( F ), so ( P ) is ( F ). But in the problem statement, it says "intersects its axis at point ( P )", so maybe ( P ) is indeed ( F ). Hmm, but then ( |AP| ) would be the distance from ( A ) to ( F ), and ( |AB| ) is the distance between ( A ) and ( B ). The problem is asking for the ratio ( frac{|AB|}{|AP|} ).But let me make sure. Maybe I misread the problem. It says the line passes through ( F ) with a slope angle of ( 60^circ ), intersects the parabola at ( A ) and ( B ), and intersects its axis at ( P ). So, perhaps ( P ) is not ( F ). Wait, if the line passes through ( F ), then when it intersects the axis (x-axis), it must pass through ( F ), so ( P ) is ( F ). That seems correct.But then, in the initial solution provided, they found ( |AP| = 4p ) and ( |AB| = frac{7}{3}p ), leading to the ratio ( frac{7}{12} ). Let me try to go through the steps myself to see if I can replicate this.First, let's write down the equation of the parabola: ( y^2 = 2px ). The focus is at ( ( frac{p}{2}, 0 ) ). The line passing through ( F ) with a slope of ( sqrt{3} ) has the equation ( y = sqrt{3}(x - frac{p}{2}) ).To find the points of intersection ( A ) and ( B ), we can substitute ( y ) from the line equation into the parabola equation. So, substituting ( y = sqrt{3}(x - frac{p}{2}) ) into ( y^2 = 2px ), we get:[ [ sqrt{3}(x - frac{p}{2}) ]^2 = 2px ][ 3(x - frac{p}{2})^2 = 2px ]Expanding the left side:[ 3(x^2 - px + frac{p^2}{4}) = 2px ][ 3x^2 - 3px + frac{3p^2}{4} = 2px ]Bring all terms to one side:[ 3x^2 - 3px - 2px + frac{3p^2}{4} = 0 ]Combine like terms:[ 3x^2 - 5px + frac{3p^2}{4} = 0 ]Multiply both sides by 4 to eliminate the fraction:[ 12x^2 - 20px + 3p^2 = 0 ]So, the quadratic equation in terms of ( x ) is ( 12x^2 - 20px + 3p^2 = 0 ). Let's solve for ( x ) using the quadratic formula:[ x = frac{20p pm sqrt{(20p)^2 - 4 cdot 12 cdot 3p^2}}{2 cdot 12} ]Calculate the discriminant:[ (20p)^2 - 4 cdot 12 cdot 3p^2 = 400p^2 - 144p^2 = 256p^2 ]So, the square root of the discriminant is ( 16p ).Thus, the solutions are:[ x = frac{20p pm 16p}{24} ]Calculating both possibilities:1. ( x = frac{20p + 16p}{24} = frac{36p}{24} = frac{3p}{2} )2. ( x = frac{20p - 16p}{24} = frac{4p}{24} = frac{p}{6} )So, the x-coordinates of points ( A ) and ( B ) are ( frac{3p}{2} ) and ( frac{p}{6} ), respectively. Let's find their corresponding y-coordinates using the line equation ( y = sqrt{3}(x - frac{p}{2}) ).For ( x = frac{3p}{2} ):[ y = sqrt{3}left( frac{3p}{2} - frac{p}{2} right) = sqrt{3}(p) = psqrt{3} ]So, point ( A ) is ( left( frac{3p}{2}, psqrt{3} right) ).For ( x = frac{p}{6} ):[ y = sqrt{3}left( frac{p}{6} - frac{p}{2} right) = sqrt{3}left( -frac{p}{3} right) = -frac{psqrt{3}}{3} ]So, point ( B ) is ( left( frac{p}{6}, -frac{psqrt{3}}{3} right) ).Now, we need to find the distances ( |AB| ) and ( |AP| ).First, let's find ( |AB| ). Using the distance formula between points ( A ) and ( B ):[ |AB| = sqrt{ left( frac{3p}{2} - frac{p}{6} right)^2 + left( psqrt{3} - left( -frac{psqrt{3}}{3} right) right)^2 } ]Simplify the differences:For the x-coordinates:[ frac{3p}{2} - frac{p}{6} = frac{9p}{6} - frac{p}{6} = frac{8p}{6} = frac{4p}{3} ]For the y-coordinates:[ psqrt{3} - left( -frac{psqrt{3}}{3} right) = psqrt{3} + frac{psqrt{3}}{3} = frac{4psqrt{3}}{3} ]Now, plug these into the distance formula:[ |AB| = sqrt{ left( frac{4p}{3} right)^2 + left( frac{4psqrt{3}}{3} right)^2 } ][ = sqrt{ frac{16p^2}{9} + frac{16 cdot 3 p^2}{9} } ][ = sqrt{ frac{16p^2}{9} + frac{48p^2}{9} } ][ = sqrt{ frac{64p^2}{9} } ][ = frac{8p}{3} ]So, ( |AB| = frac{8p}{3} ).Next, let's find ( |AP| ). Point ( P ) is the intersection of line ( l ) with the axis of the parabola, which is the x-axis. As we determined earlier, this occurs at ( F ), which is ( ( frac{p}{2}, 0 ) ).Wait, hold on. Earlier, I thought ( P ) was ( F ), but in the initial solution, they mentioned ( |AP| = 4p ). Let me check if this is correct.Point ( A ) is ( left( frac{3p}{2}, psqrt{3} right) ) and point ( P ) is ( ( frac{p}{2}, 0 ) ). So, the distance ( |AP| ) is:[ |AP| = sqrt{ left( frac{3p}{2} - frac{p}{2} right)^2 + left( psqrt{3} - 0 right)^2 } ][ = sqrt{ left( p right)^2 + left( psqrt{3} right)^2 } ][ = sqrt{ p^2 + 3p^2 } ][ = sqrt{4p^2} ][ = 2p ]Wait, that's different from the initial solution which said ( |AP| = 4p ). Did I make a mistake?Wait, maybe I misread the problem. It says the line intersects the axis at point ( P ). If the axis is the x-axis, then yes, ( P ) is ( ( frac{p}{2}, 0 ) ). But the initial solution had ( |AP| = 4p ), which doesn't match my calculation here. So, perhaps I need to double-check.Wait, let me recalculate ( |AP| ):Point ( A ): ( left( frac{3p}{2}, psqrt{3} right) )Point ( P ): ( left( frac{p}{2}, 0 right) )Difference in x-coordinates: ( frac{3p}{2} - frac{p}{2} = p )Difference in y-coordinates: ( psqrt{3} - 0 = psqrt{3} )So, distance ( |AP| = sqrt{p^2 + (psqrt{3})^2} = sqrt{p^2 + 3p^2} = sqrt{4p^2} = 2p ). So, it's definitely ( 2p ), not ( 4p ). So, the initial solution might have an error here.But wait, in the initial solution, they wrote ( |AB| = x_1 + x_2 + p = frac{7}{3}p ). Hmm, that seems off because ( |AB| ) is a distance, not just the sum of x-coordinates. So, perhaps their method was incorrect.Wait, let me think. They found ( x_1 = frac{3}{2}p ) and ( x_2 = frac{1}{6}p ), which are correct. Then, they said ( |AB| = x_1 + x_2 + p ). That seems like they added the x-coordinates and then added ( p ). But why would they do that? Maybe they thought of the distance along the x-axis or something else, but in reality, the distance between two points isn't just the sum of their x-coordinates plus something else.So, in my calculation, I found ( |AB| = frac{8p}{3} ), which is approximately ( 2.666p ), and ( |AP| = 2p ). Therefore, the ratio ( frac{|AB|}{|AP|} = frac{frac{8p}{3}}{2p} = frac{8}{6} = frac{4}{3} ). But that contradicts the initial solution's ratio of ( frac{7}{12} ).Wait, maybe I made a mistake in calculating ( |AB| ). Let me go through that again.Points ( A ) and ( B ):( A ): ( left( frac{3p}{2}, psqrt{3} right) )( B ): ( left( frac{p}{6}, -frac{psqrt{3}}{3} right) )Differences:Δx = ( frac{3p}{2} - frac{p}{6} = frac{9p}{6} - frac{p}{6} = frac{8p}{6} = frac{4p}{3} )Δy = ( psqrt{3} - (-frac{psqrt{3}}{3}) = psqrt{3} + frac{psqrt{3}}{3} = frac{4psqrt{3}}{3} )So, distance ( |AB| = sqrt{ left( frac{4p}{3} right)^2 + left( frac{4psqrt{3}}{3} right)^2 } )Calculating each term:( left( frac{4p}{3} right)^2 = frac{16p^2}{9} )( left( frac{4psqrt{3}}{3} right)^2 = frac{16p^2 cdot 3}{9} = frac{48p^2}{9} )Adding them together:( frac{16p^2}{9} + frac{48p^2}{9} = frac{64p^2}{9} )Taking the square root:( sqrt{ frac{64p^2}{9} } = frac{8p}{3} )So, yes, ( |AB| = frac{8p}{3} ). That seems correct.Then, ( |AP| = 2p ), so the ratio ( frac{|AB|}{|AP|} = frac{frac{8p}{3}}{2p} = frac{8}{6} = frac{4}{3} ). So, that would be ( frac{4}{3} ).But wait, in the initial solution, they had ( |AB| = frac{7}{3}p ) and ( |AP| = 4p ), giving ( frac{7}{12} ). So, clearly, there's a discrepancy here. Maybe the initial solution was incorrect.Alternatively, perhaps I misunderstood the definition of ( P ). The problem says the line intersects the axis at point ( P ). If the axis is the x-axis, then ( P ) is ( ( frac{p}{2}, 0 ) ), which is ( F ). But maybe the axis refers to the axis of the parabola, which is the x-axis, so yes, ( P ) is ( F ).Wait, but in the initial solution, they found ( |AP| = 4p ). How did they get that? Let me see.They wrote: "Solving this equation along with the parabola equation and eliminating ( y ), we get ( 12x^2 - 20px + 3p^2 = 0 ). This gives us ( x_1x_2 = frac{p^2}{4} ), hence ( x_1 = frac{3}{2}p ) and ( x_2 = frac{1}{6}p ). Now, ( |AB| = x_1 + x_2 + p = frac{7}{3}p ), and ( |AP| = 4p )."Wait, so they added ( x_1 + x_2 + p ) to get ( |AB| ). That doesn't make sense because ( |AB| ) is a distance, not a sum of x-coordinates. Also, they said ( |AP| = 4p ), which is inconsistent with my calculation of ( 2p ).So, perhaps the initial solution had a mistake in calculating ( |AB| ) and ( |AP| ). Let me try to figure out where they went wrong.First, solving the quadratic equation ( 12x^2 - 20px + 3p^2 = 0 ), the roots are ( x_1 = frac{3p}{2} ) and ( x_2 = frac{p}{6} ). So, that's correct.Then, they said ( |AB| = x_1 + x_2 + p ). Wait, why would they add ( p )? The x-coordinates of ( A ) and ( B ) are ( frac{3p}{2} ) and ( frac{p}{6} ). The distance between ( A ) and ( B ) isn't just the sum of their x-coordinates plus ( p ). That seems like an incorrect approach.Similarly, for ( |AP| ), they might have miscalculated. If ( P ) is ( F ), which is ( ( frac{p}{2}, 0 ) ), then ( |AP| ) should be the distance between ( A ) and ( F ), which is ( 2p ), as I calculated.Alternatively, maybe ( P ) is not ( F ). Let me re-examine the problem statement: "a line ( l ) passing through ( F ) with a slope angle of ( 60^circ ) intersects the parabola ( C ) at points ( A ) and ( B ) in the first and fourth quadrants, respectively, and intersects its axis at point ( P )."So, the line passes through ( F ) and intersects the axis (x-axis) at ( P ). Since the line passes through ( F ), which is on the x-axis, ( P ) must coincide with ( F ). Therefore, ( P ) is ( F ).But in that case, ( |AP| ) is the distance from ( A ) to ( F ), which is ( 2p ), as I found. So, the initial solution's ( |AP| = 4p ) must be incorrect.Alternatively, perhaps the problem defines the axis differently. Wait, the axis of the parabola ( y^2 = 2px ) is indeed the x-axis. So, the line intersects the x-axis at ( P ), which is ( F ).Wait, maybe the initial solution considered the axis as the line perpendicular to the parabola's axis, but that doesn't make sense. The axis of a parabola is its line of symmetry, which in this case is the x-axis.Wait, perhaps I made a mistake in calculating ( |AP| ). Let me double-check.Point ( A ): ( left( frac{3p}{2}, psqrt{3} right) )Point ( P ): ( left( frac{p}{2}, 0 right) )Difference in x: ( frac{3p}{2} - frac{p}{2} = p )Difference in y: ( psqrt{3} - 0 = psqrt{3} )Distance ( |AP| = sqrt{p^2 + (psqrt{3})^2} = sqrt{p^2 + 3p^2} = sqrt{4p^2} = 2p ). So, that's correct.So, the initial solution's ( |AP| = 4p ) is wrong. Therefore, their ratio ( frac{7}{12} ) is incorrect.Wait, but in my calculation, ( |AB| = frac{8p}{3} ) and ( |AP| = 2p ), so the ratio is ( frac{8}{3} / 2 = frac{4}{3} ). But that contradicts the initial solution. So, perhaps I need to think differently.Wait, maybe the initial solution considered the length along the line ( l ) instead of the straight-line distance. Let me see.If we consider the distance from ( A ) to ( P ) along the line ( l ), which has a slope of ( sqrt{3} ), so the angle is ( 60^circ ). The distance from ( A ) to ( P ) along the line would be the length of the line segment ( AP ), which we calculated as ( 2p ). But if we consider the distance along the x-axis, that would be different, but the problem just says ( |AP| ), which is the straight-line distance.Alternatively, maybe the initial solution used parametric equations or vector methods. Let me try that approach.Parametrize the line ( l ) as follows:Since it passes through ( F ) ( ( frac{p}{2}, 0 ) ) and has a slope of ( sqrt{3} ), we can write the parametric equations as:( x = frac{p}{2} + t )( y = 0 + sqrt{3}t )Where ( t ) is a parameter.Now, substitute these into the parabola equation ( y^2 = 2px ):( ( sqrt{3}t )^2 = 2p left( frac{p}{2} + t right ) )( 3t^2 = p^2 + 2pt )Bring all terms to one side:( 3t^2 - 2pt - p^2 = 0 )Solving for ( t ):[ t = frac{2p pm sqrt{(2p)^2 + 12p^2}}{6} ][ = frac{2p pm sqrt{16p^2}}{6} ][ = frac{2p pm 4p}{6} ]So, the solutions are:1. ( t = frac{2p + 4p}{6} = frac{6p}{6} = p )2. ( t = frac{2p - 4p}{6} = frac{-2p}{6} = -frac{p}{3} )So, the parameter values are ( t = p ) and ( t = -frac{p}{3} ).Therefore, the points ( A ) and ( B ) correspond to these ( t ) values.For ( t = p ):( x = frac{p}{2} + p = frac{3p}{2} )( y = sqrt{3}p )So, point ( A ) is ( left( frac{3p}{2}, sqrt{3}p right) )For ( t = -frac{p}{3} ):( x = frac{p}{2} - frac{p}{3} = frac{3p - 2p}{6} = frac{p}{6} )( y = sqrt{3}(-frac{p}{3}) = -frac{sqrt{3}p}{3} )So, point ( B ) is ( left( frac{p}{6}, -frac{sqrt{3}p}{3} right) )Now, using the parameter ( t ), the distance from ( A ) to ( P ) can be found by considering the parameter difference. Since ( P ) is at ( t = 0 ) (because when ( t = 0 ), ( x = frac{p}{2} ), ( y = 0 )), the distance from ( A ) (at ( t = p )) to ( P ) (at ( t = 0 )) along the line ( l ) is the length corresponding to ( t = p ).But wait, the distance along the line isn't just ( t ), because the parameter ( t ) is scaled by the direction vector. The direction vector of the line ( l ) is ( (1, sqrt{3}) ), which has a magnitude of ( sqrt{1 + 3} = 2 ). So, each unit of ( t ) corresponds to a distance of 2 units along the line.Therefore, the distance from ( A ) to ( P ) along the line ( l ) is ( |t| times text{magnitude} = p times 2 = 2p ). But this is the same as the straight-line distance ( |AP| ), which we already calculated as ( 2p ).Similarly, the distance from ( A ) to ( B ) along the line ( l ) would be the difference in their ( t ) values times the magnitude. The ( t ) values are ( p ) and ( -frac{p}{3} ), so the difference is ( p - (-frac{p}{3}) = frac{4p}{3} ). Therefore, the distance along the line is ( frac{4p}{3} times 2 = frac{8p}{3} ), which matches our earlier calculation of ( |AB| = frac{8p}{3} ).So, whether we calculate the straight-line distance or the distance along the line ( l ), we get the same numerical value for ( |AB| ) and ( |AP| ). Therefore, the ratio ( frac{|AB|}{|AP|} = frac{frac{8p}{3}}{2p} = frac{4}{3} ).But wait, the initial solution had a different ratio. Maybe the problem is asking for something else. Let me re-read the problem statement."Find the value of ( frac{|AB|}{|AP|} )."So, it's the ratio of the distance between ( A ) and ( B ) to the distance between ( A ) and ( P ). As we calculated, ( |AB| = frac{8p}{3} ) and ( |AP| = 2p ), so the ratio is ( frac{8}{6} = frac{4}{3} ).But the initial solution had ( frac{7}{12} ). So, perhaps there's a misunderstanding in the problem's interpretation.Wait, maybe the initial solution considered ( |AB| ) as the sum of the roots of the quadratic equation, which are ( x_1 + x_2 = frac{3p}{2} + frac{p}{6} = frac{10p}{6} = frac{5p}{3} ), and then added ( p ) for some reason, getting ( frac{5p}{3} + p = frac{8p}{3} ), which is actually correct. But then they said ( |AB| = frac{7}{3}p ), which is inconsistent.Wait, no, in the initial solution, they wrote ( |AB| = x_1 + x_2 + p = frac{7}{3}p ). But ( x_1 + x_2 = frac{3p}{2} + frac{p}{6} = frac{9p + p}{6} = frac{10p}{6} = frac{5p}{3} ). So, adding ( p ) gives ( frac{5p}{3} + p = frac{8p}{3} ), which is correct. But they wrote ( frac{7}{3}p ), which is incorrect. So, that was a mistake in their calculation.Similarly, for ( |AP| ), they said it was ( 4p ), but it's actually ( 2p ). So, their ratio ( frac{7}{12} ) is wrong.Therefore, the correct ratio should be ( frac{4}{3} ).But wait, let me think again. Maybe the problem is asking for the ratio in terms of the parameter ( t ). Since ( |AB| ) corresponds to ( frac{8p}{3} ) and ( |AP| ) corresponds to ( 2p ), the ratio is ( frac{4}{3} ).Alternatively, perhaps the initial solution considered the length from ( A ) to ( P ) as the entire length from ( A ) to ( B ) plus something else, but that doesn't make sense.Wait, another thought: maybe the initial solution considered ( |AP| ) as the distance from ( A ) to ( P ) along the x-axis, which would be different. Let me check.If ( P ) is ( ( frac{p}{2}, 0 ) ), then the horizontal distance from ( A ) to ( P ) is ( frac{3p}{2} - frac{p}{2} = p ). But that's not the straight-line distance, just the x-component. So, if they considered only the x-component, they might have thought ( |AP| = p ), but that's not the case.Alternatively, maybe they considered the distance from ( A ) to ( P ) along the line ( l ), which is ( 2p ), as we saw earlier. But they said ( |AP| = 4p ), which is double that.Wait, maybe they made a mistake in calculating the parameter ( t ). In the parametric approach, the distance from ( A ) to ( P ) is ( 2p ), but if they thought the parameter ( t ) was the actual distance, they might have miscalculated.Wait, in the parametric equations, ( t ) is not the distance, but the parameter. The actual distance is ( t times ) the magnitude of the direction vector. Since the direction vector is ( (1, sqrt{3}) ), its magnitude is 2, so the distance is ( 2t ). For ( t = p ), the distance is ( 2p ), which is correct. So, if they thought ( t ) was the distance, they might have said ( |AP| = p ), but that's not the case.Alternatively, perhaps they considered the distance from ( A ) to ( P ) as the sum of ( |AF| ) and ( |FP| ), but ( P ) is ( F ), so that would just be ( |AF| ), which is ( 2p ).Wait, another idea: maybe they considered ( |AP| ) as the distance from ( A ) to ( P ) along the x-axis, which is ( p ), and then multiplied by 2 for some reason, getting ( 2p times 2 = 4p ). But that's not correct because ( |AP| ) is the straight-line distance, not the sum of x and y components.In any case, based on my calculations, the correct ratio is ( frac{4}{3} ). However, the initial solution had ( frac{7}{12} ), which seems to be incorrect due to miscalculations in ( |AB| ) and ( |AP| ).Wait, perhaps I made a mistake in calculating ( |AB| ). Let me try another approach. Since ( A ) and ( B ) are points on the parabola, maybe there's a property or formula that relates the distances between points on a parabola intersected by a line through the focus.I recall that for a parabola, the focal length is ( frac{p}{2} ), and there are properties related to the distances from the focus to points on the parabola, but I'm not sure if that directly helps here.Alternatively, maybe using the parametric form of the parabola. The standard parametric equations for ( y^2 = 4ax ) are ( x = at^2 ), ( y = 2at ). But our parabola is ( y^2 = 2px ), which can be written as ( y^2 = 4a x ) with ( 4a = 2p ), so ( a = frac{p}{2} ). Therefore, the parametric equations would be ( x = frac{p}{2} t^2 ), ( y = p t ).So, points ( A ) and ( B ) can be represented as ( ( frac{p}{2} t_1^2, p t_1 ) ) and ( ( frac{p}{2} t_2^2, p t_2 ) ).The line passing through the focus ( ( frac{p}{2}, 0 ) ) with slope ( sqrt{3} ) can be written as ( y = sqrt{3}(x - frac{p}{2}) ).Substituting the parametric coordinates into the line equation:( p t = sqrt{3} left( frac{p}{2} t^2 - frac{p}{2} right ) )Simplify:( p t = sqrt{3} cdot frac{p}{2} (t^2 - 1) )Divide both sides by ( p ):( t = frac{sqrt{3}}{2} (t^2 - 1) )Multiply both sides by 2:( 2t = sqrt{3} (t^2 - 1) )Bring all terms to one side:( sqrt{3} t^2 - 2t - sqrt{3} = 0 )This is a quadratic in ( t ). Let's solve for ( t ):[ t = frac{2 pm sqrt{4 + 4 cdot 3}}{2sqrt{3}} ][ = frac{2 pm sqrt{16}}{2sqrt{3}} ][ = frac{2 pm 4}{2sqrt{3}} ]So, the solutions are:1. ( t = frac{6}{2sqrt{3}} = frac{3}{sqrt{3}} = sqrt{3} )2. ( t = frac{-2}{2sqrt{3}} = -frac{1}{sqrt{3}} )Therefore, the parameters are ( t_1 = sqrt{3} ) and ( t_2 = -frac{1}{sqrt{3}} ).So, points ( A ) and ( B ) are:( A ): ( left( frac{p}{2} (sqrt{3})^2, p sqrt{3} right ) = left( frac{p}{2} cdot 3, psqrt{3} right ) = left( frac{3p}{2}, psqrt{3} right ) )( B ): ( left( frac{p}{2} left( -frac{1}{sqrt{3}} right )^2, p left( -frac{1}{sqrt{3}} right ) right ) = left( frac{p}{2} cdot frac{1}{3}, -frac{p}{sqrt{3}} right ) = left( frac{p}{6}, -frac{psqrt{3}}{3} right ) )Which matches our earlier calculations.Now, using the parametric form, the distance ( |AB| ) can be calculated using the distance formula, which we already did, giving ( frac{8p}{3} ).Alternatively, using the parametric properties, the distance between two points on a parabola can sometimes be expressed in terms of their parameters. For a parabola ( y^2 = 4ax ), the distance between two points with parameters ( t_1 ) and ( t_2 ) is given by ( sqrt{(at_1^2 - at_2^2)^2 + (2at_1 - 2at_2)^2} ). But in our case, the parabola is ( y^2 = 2px ), so ( a = frac{p}{2} ).So, the distance ( |AB| ) would be:[ sqrt{ left( frac{p}{2} t_1^2 - frac{p}{2} t_2^2 right )^2 + left( p t_1 - p t_2 right )^2 } ][ = sqrt{ left( frac{p}{2} (t_1^2 - t_2^2) right )^2 + left( p(t_1 - t_2) right )^2 } ]Factor out ( p^2 ):[ = p sqrt{ left( frac{1}{2} (t_1^2 - t_2^2) right )^2 + (t_1 - t_2)^2 } ]Plugging in ( t_1 = sqrt{3} ) and ( t_2 = -frac{1}{sqrt{3}} ):First, calculate ( t_1^2 - t_2^2 ):( (sqrt{3})^2 - left( -frac{1}{sqrt{3}} right )^2 = 3 - frac{1}{3} = frac{8}{3} )Then, ( t_1 - t_2 = sqrt{3} - left( -frac{1}{sqrt{3}} right ) = sqrt{3} + frac{1}{sqrt{3}} = frac{3 + 1}{sqrt{3}} = frac{4}{sqrt{3}} )Now, plug these into the distance formula:[ |AB| = p sqrt{ left( frac{1}{2} cdot frac{8}{3} right )^2 + left( frac{4}{sqrt{3}} right )^2 } ][ = p sqrt{ left( frac{4}{3} right )^2 + left( frac{16}{3} right ) } ][ = p sqrt{ frac{16}{9} + frac{16}{3} } ]Convert ( frac{16}{3} ) to ninths:[ = p sqrt{ frac{16}{9} + frac{48}{9} } ][ = p sqrt{ frac{64}{9} } ][ = p cdot frac{8}{3} ][ = frac{8p}{3} ]So, that confirms ( |AB| = frac{8p}{3} ).Now, for ( |AP| ), using the parametric form, point ( P ) is ( ( frac{p}{2}, 0 ) ), which corresponds to the parameter ( t = 0 ). So, the distance from ( A ) (at ( t = sqrt{3} )) to ( P ) (at ( t = 0 )) can be calculated using the distance formula:[ |AP| = sqrt{ left( frac{3p}{2} - frac{p}{2} right )^2 + left( psqrt{3} - 0 right )^2 } ][ = sqrt{ (p)^2 + (psqrt{3})^2 } ][ = sqrt{ p^2 + 3p^2 } ][ = sqrt{4p^2} ][ = 2p ]So, that's consistent.Therefore, the ratio ( frac{|AB|}{|AP|} = frac{frac{8p}{3}}{2p} = frac{4}{3} ).But wait, the initial solution had ( frac{7}{12} ). So, unless there's a different interpretation of the problem, I think my calculation is correct, and the initial solution had errors in calculating ( |AB| ) and ( |AP| ).Alternatively, maybe the problem was asking for something else, like the ratio of the lengths along the line ( l ) or something else, but the problem statement clearly asks for ( frac{|AB|}{|AP|} ), which are straight-line distances.Therefore, after carefully going through the problem and recalculating, I conclude that the correct ratio is ( frac{4}{3} ).But wait, let me check once more. Maybe I made a mistake in the parametric approach.In the parametric approach, the distance between two points on the parabola is given by ( sqrt{(x_1 - x_2)^2 + (y_1 - y_2)^2} ), which we did, and it gave ( frac{8p}{3} ). The distance from ( A ) to ( P ) is ( 2p ). So, the ratio is ( frac{4}{3} ).Alternatively, maybe the problem is asking for the ratio of the lengths of segments ( AB ) and ( AP ) along the line ( l ), which would be the same as the ratio of their parameter differences times the direction vector's magnitude.Since ( |AB| ) corresponds to ( t_1 - t_2 = sqrt{3} - (-frac{1}{sqrt{3}}) = sqrt{3} + frac{1}{sqrt{3}} = frac{4}{sqrt{3}} ), and the direction vector's magnitude is 2, so the distance is ( frac{4}{sqrt{3}} times 2 = frac{8}{sqrt{3}} ). Wait, that's different from our earlier calculation. Wait, no, actually, the parameter ( t ) is related to the direction vector, so the distance is ( |t_1 - t_2| times text{magnitude} ).Wait, in the parametric equations, the direction vector is ( (1, sqrt{3}) ), which has a magnitude of 2. So, each unit of ( t ) corresponds to 2 units of distance. Therefore, the distance between ( A ) and ( B ) is ( |t_1 - t_2| times 2 = left| sqrt{3} - (-frac{1}{sqrt{3}}) right| times 2 = left| sqrt{3} + frac{1}{sqrt{3}} right| times 2 = left( frac{4}{sqrt{3}} right ) times 2 = frac{8}{sqrt{3}} ). But this is not the same as our earlier calculation of ( frac{8p}{3} ). Wait, that can't be, because ( p ) is a parameter in the equation.Wait, no, in the parametric approach, the distance is ( |t_1 - t_2| times text{magnitude} ), but the magnitude is 2, so:Distance ( |AB| = |t_1 - t_2| times 2 = left| sqrt{3} - (-frac{1}{sqrt{3}}) right| times 2 = left( sqrt{3} + frac{1}{sqrt{3}} right ) times 2 = left( frac{3 + 1}{sqrt{3}} right ) times 2 = frac{4}{sqrt{3}} times 2 = frac{8}{sqrt{3}} ).But this is in terms of the parameter scaling, not in terms of ( p ). Wait, no, actually, in the parametric equations, ( x = frac{p}{2} t^2 ) and ( y = p t ), so the actual distance between points ( A ) and ( B ) is:[ sqrt{ left( frac{p}{2} t_1^2 - frac{p}{2} t_2^2 right )^2 + left( p t_1 - p t_2 right )^2 } ]Which we already calculated as ( frac{8p}{3} ).So, the parametric approach confirms that ( |AB| = frac{8p}{3} ) and ( |AP| = 2p ), leading to the ratio ( frac{4}{3} ).Therefore, I'm confident that the correct ratio is ( frac{4}{3} ), and the initial solution had errors in calculating ( |AB| ) and ( |AP| ).But wait, the initial solution had ( |AB| = frac{7}{3}p ) and ( |AP| = 4p ), giving ( frac{7}{12} ). So, perhaps they made a mistake in solving the quadratic equation.Let me re-examine their quadratic equation:They wrote: "Solving this equation along with the parabola equation and eliminating ( y ), we get ( 12x^2 - 20px + 3p^2 = 0 ). This gives us ( x_1x_2 = frac{p^2}{4} ), hence ( x_1 = frac{3}{2}p ) and ( x_2 = frac{1}{6}p )."Wait, the quadratic equation is ( 12x^2 - 20px + 3p^2 = 0 ). The product of the roots is ( frac{3p^2}{12} = frac{p^2}{4} ), which is correct. So, ( x_1x_2 = frac{p^2}{4} ). But from the quadratic, the sum of the roots is ( frac{20p}{12} = frac{5p}{3} ). So, ( x_1 + x_2 = frac{5p}{3} ).But in their solution, they said ( x_1 = frac{3}{2}p ) and ( x_2 = frac{1}{6}p ). Let's check if these satisfy ( x_1 + x_2 = frac{5p}{3} ):( frac{3}{2}p + frac{1}{6}p = frac{9}{6}p + frac{1}{6}p = frac{10}{6}p = frac{5}{3}p ). Yes, that's correct.So, their roots are correct. Then, they said ( |AB| = x_1 + x_2 + p = frac{7}{3}p ). Wait, ( x_1 + x_2 = frac{5p}{3} ), so adding ( p ) gives ( frac{5p}{3} + p = frac{8p}{3} ), which is correct. But they wrote ( frac{7}{3}p ), which is incorrect. So, that was a typo or miscalculation.Similarly, for ( |AP| ), they said it was ( 4p ), but it's actually ( 2p ). So, their ratio was based on incorrect values.Therefore, the correct ratio is ( frac{4}{3} ).But wait, let me think again. Maybe the problem is asking for the ratio in terms of the parameter ( t ), but no, the problem clearly asks for the ratio of the distances ( |AB| ) and ( |AP| ).Alternatively, perhaps the initial solution considered the distance from ( A ) to ( P ) as the sum of ( |AF| ) and ( |FP| ), but since ( P = F ), that would just be ( |AF| ), which is ( 2p ).Wait, another thought: maybe the initial solution considered the distance from ( A ) to ( P ) along the x-axis, which is ( p ), and then multiplied by 2 for some reason, getting ( 2p times 2 = 4p ). But that's not correct because ( |AP| ) is the straight-line distance, not the sum of x and y components.In conclusion, after carefully recalculating and verifying through multiple methods (direct distance formula, parametric equations), I'm confident that the correct ratio is ( frac{4}{3} ). The initial solution had errors in calculating ( |AB| ) and ( |AP| ), leading to an incorrect ratio of ( frac{7}{12} ).Therefore, the correct value of ( frac{|AB|}{|AP|} ) is ( frac{4}{3} ).

💡Okay, so I have this function f(x) = a x² - 2x + ln x, where a is not zero and is a real number. The problem has two parts: first, I need to determine the monotonicity of f(x), which means figuring out where it's increasing or decreasing. Second, if the function has two extreme points x₁ and x₂, I need to prove that f(x₁) + f(x₂) is less than -3.Starting with part (1). To find the monotonicity, I remember that I need to look at the first derivative of f(x). So, let's compute f'(x).f(x) = a x² - 2x + ln xTaking the derivative term by term:- The derivative of a x² is 2a x.- The derivative of -2x is -2.- The derivative of ln x is 1/x.So, putting it all together:f'(x) = 2a x - 2 + 1/xHmm, that's f'(x). To make it easier to analyze, maybe I can combine the terms over a common denominator. Let's see:f'(x) = (2a x² - 2x + 1) / xYes, that looks right. So, f'(x) is equal to (2a x² - 2x + 1) divided by x. Now, since the denominator x is always positive for x > 0 (because ln x is only defined for x > 0), the sign of f'(x) depends on the numerator: 2a x² - 2x + 1.Let me denote the numerator as g(x) = 2a x² - 2x + 1. So, f'(x) = g(x)/x. Therefore, the sign of f'(x) is the same as the sign of g(x).To find where f'(x) is positive or negative, I need to analyze g(x). Since g(x) is a quadratic function in terms of x, I can find its discriminant to see how many real roots it has.The discriminant D of a quadratic ax² + bx + c is D = b² - 4ac. For g(x), a is 2a, b is -2, and c is 1. So,D = (-2)² - 4*(2a)*1 = 4 - 8aSo, D = 4 - 8a.Now, the number of real roots depends on the discriminant:- If D > 0, two distinct real roots.- If D = 0, one real root (a repeated root).- If D < 0, no real roots.So, D = 4 - 8a. Let's solve for when D is positive, zero, or negative.Case 1: D > 04 - 8a > 0-8a > -4Divide both sides by -8 (remembering to reverse the inequality):a < 1/2Case 2: D = 04 - 8a = 0-8a = -4a = 1/2Case 3: D < 04 - 8a < 0-8a < -4a > 1/2So, depending on the value of a, the quadratic g(x) will have two, one, or no real roots.Let's analyze each case.Case 1: a ≥ 1/2If a is greater than or equal to 1/2, then D ≤ 0. So, g(x) has no real roots or one real root.If D = 0 (a = 1/2), then g(x) has a repeated root. Let's find that root.g(x) = 2*(1/2)*x² - 2x + 1 = x² - 2x + 1 = (x - 1)^2So, the root is x = 1, and it's a double root.Since a is positive (a = 1/2), the parabola opens upwards. Therefore, g(x) is always non-negative because it has a double root and opens upwards. So, g(x) ≥ 0 for all x.Therefore, f'(x) = g(x)/x ≥ 0 for all x > 0. So, f(x) is increasing on (0, ∞).Case 2: a < 1/2Now, if a is less than 1/2, D > 0, so g(x) has two distinct real roots. Let's find these roots.Using the quadratic formula:x = [2 ± sqrt(4 - 8a)] / (2*2a) = [2 ± sqrt(4 - 8a)] / (4a) = [1 ± sqrt(1 - 2a)] / (2a)So, the two roots are:x₁ = [1 - sqrt(1 - 2a)] / (2a)x₂ = [1 + sqrt(1 - 2a)] / (2a)Now, we need to analyze the behavior of g(x) based on the value of a.Subcase 2a: 0 < a < 1/2Here, a is positive but less than 1/2. So, sqrt(1 - 2a) is a real number because 1 - 2a > 0.Let's compute x₁ and x₂.Since a is positive, both denominators are positive.Compute x₁:x₁ = [1 - sqrt(1 - 2a)] / (2a)Since sqrt(1 - 2a) < 1 (because 1 - 2a < 1), the numerator is positive: 1 - sqrt(1 - 2a) > 0.Therefore, x₁ is positive.Similarly, x₂ = [1 + sqrt(1 - 2a)] / (2a) is also positive.So, both roots are positive. Therefore, the quadratic g(x) opens upwards (since the coefficient of x² is 2a > 0) and has two positive roots.Therefore, the graph of g(x) is a parabola opening upwards, crossing the x-axis at x₁ and x₂.Therefore, g(x) is positive outside the interval (x₁, x₂) and negative inside (x₁, x₂).Therefore, f'(x) is positive when x < x₁ or x > x₂, and negative when x₁ < x < x₂.Therefore, f(x) is increasing on (0, x₁), decreasing on (x₁, x₂), and increasing again on (x₂, ∞).Subcase 2b: a < 0Here, a is negative. So, sqrt(1 - 2a) is still real because 1 - 2a > 1 (since a is negative, subtracting a negative makes it larger). So, sqrt(1 - 2a) is greater than 1.Compute x₁ and x₂.x₁ = [1 - sqrt(1 - 2a)] / (2a)x₂ = [1 + sqrt(1 - 2a)] / (2a)Since a is negative, the denominators are negative.Compute x₁:Numerator: 1 - sqrt(1 - 2a). Since sqrt(1 - 2a) > 1, 1 - sqrt(1 - 2a) is negative.Denominator: 2a is negative.So, x₁ = (negative)/(negative) = positive.Similarly, x₂:Numerator: 1 + sqrt(1 - 2a). Since sqrt(1 - 2a) > 1, this is positive.Denominator: 2a is negative.So, x₂ = (positive)/(negative) = negative.But x₂ is negative, which is outside the domain of f(x) (since x > 0). Therefore, only x₁ is in the domain.So, for a < 0, g(x) has two roots, but only x₁ is positive.Since a is negative, the parabola g(x) opens downward (because the coefficient of x² is 2a < 0).So, the parabola opens downward, crosses the x-axis at x₁ (positive) and x₂ (negative). Therefore, for x > 0, g(x) is positive before x₁ and negative after x₁.Wait, let's think carefully.Since the parabola opens downward, it will be positive between its two roots and negative outside. But since one root is negative and the other is positive, for x > 0, the function g(x) is positive between x = 0 and x = x₁, and negative for x > x₁.Wait, but actually, since the parabola opens downward, it will be positive between the two roots. But since one root is negative and the other is positive, for x > 0, the function is positive from x = 0 up to x = x₁, and then negative beyond x₁.But wait, at x = 0, g(x) = 1, which is positive. So, as x increases from 0, g(x) remains positive until x = x₁, then becomes negative beyond x₁.Therefore, f'(x) = g(x)/x is positive when x < x₁ and negative when x > x₁.Therefore, f(x) is increasing on (0, x₁) and decreasing on (x₁, ∞).So, summarizing part (1):- If a ≥ 1/2, f(x) is increasing on (0, ∞).- If 0 < a < 1/2, f(x) is increasing on (0, x₁), decreasing on (x₁, x₂), and increasing again on (x₂, ∞).- If a < 0, f(x) is increasing on (0, x₁) and decreasing on (x₁, ∞).Okay, that seems to cover all cases for part (1).Moving on to part (2). We are told that f(x) has two extreme points x₁ and x₂, which implies that the function has two critical points, meaning that f'(x) has two zeros. From part (1), this happens when 0 < a < 1/2. So, in this case, f(x) has a local maximum at x₁ and a local minimum at x₂, or vice versa? Wait, let's think.From part (1), when 0 < a < 1/2, f(x) is increasing on (0, x₁), decreasing on (x₁, x₂), and increasing again on (x₂, ∞). So, x₁ is a local maximum, and x₂ is a local minimum.Therefore, x₁ is a local maximum, and x₂ is a local minimum.We need to prove that f(x₁) + f(x₂) < -3.So, first, let's find expressions for f(x₁) and f(x₂). Since x₁ and x₂ are critical points, f'(x₁) = 0 and f'(x₂) = 0.From f'(x) = (2a x² - 2x + 1)/x = 0, so 2a x² - 2x + 1 = 0.So, for x = x₁ and x = x₂, we have:2a x₁² - 2x₁ + 1 = 02a x₂² - 2x₂ + 1 = 0We can solve for a in terms of x:From the first equation:2a x₁² = 2x₁ - 1So, a = (2x₁ - 1)/(2x₁²)Similarly, a = (2x₂ - 1)/(2x₂²)Since both equal a, we can set them equal:(2x₁ - 1)/(2x₁²) = (2x₂ - 1)/(2x₂²)Cross-multiplying:(2x₁ - 1) * 2x₂² = (2x₂ - 1) * 2x₁²Simplify:(2x₁ - 1) x₂² = (2x₂ - 1) x₁²Expanding both sides:2x₁ x₂² - x₂² = 2x₂ x₁² - x₁²Bring all terms to one side:2x₁ x₂² - x₂² - 2x₂ x₁² + x₁² = 0Factor terms:2x₁ x₂² - 2x₂ x₁² - x₂² + x₁² = 0Factor 2x₁ x₂² - 2x₂ x₁² as 2x₁ x₂ (x₂ - x₁)And factor -x₂² + x₁² as -(x₂² - x₁²) = -(x₂ - x₁)(x₂ + x₁)So, putting it together:2x₁ x₂ (x₂ - x₁) - (x₂ - x₁)(x₂ + x₁) = 0Factor out (x₂ - x₁):(x₂ - x₁)(2x₁ x₂ - (x₂ + x₁)) = 0Since x₂ ≠ x₁ (they are distinct roots), we have:2x₁ x₂ - (x₂ + x₁) = 0So,2x₁ x₂ = x₁ + x₂Let me denote S = x₁ + x₂ and P = x₁ x₂.From the quadratic equation g(x) = 2a x² - 2x + 1 = 0, the sum of roots S = (2)/(2a) = 1/aAnd the product P = (1)/(2a)So, S = 1/a and P = 1/(2a)From the equation above, 2P = SSo,2*(1/(2a)) = 1/aWhich simplifies to 1/a = 1/a, which is consistent. So, that checks out.Now, we need to find f(x₁) + f(x₂).Let's compute f(x₁) and f(x₂).f(x) = a x² - 2x + ln xSo,f(x₁) = a x₁² - 2x₁ + ln x₁f(x₂) = a x₂² - 2x₂ + ln x₂So, f(x₁) + f(x₂) = a(x₁² + x₂²) - 2(x₁ + x₂) + (ln x₁ + ln x₂)Simplify:= a(x₁² + x₂²) - 2S + ln(x₁ x₂)We know that S = x₁ + x₂ = 1/a and P = x₁ x₂ = 1/(2a)Also, x₁² + x₂² = (x₁ + x₂)^2 - 2x₁ x₂ = S² - 2PSo,x₁² + x₂² = (1/a)^2 - 2*(1/(2a)) = 1/a² - 1/aTherefore,f(x₁) + f(x₂) = a*(1/a² - 1/a) - 2*(1/a) + ln(1/(2a))Simplify term by term:First term: a*(1/a² - 1/a) = a*(1/a²) - a*(1/a) = 1/a - 1Second term: -2*(1/a) = -2/aThird term: ln(1/(2a)) = ln(1) - ln(2a) = 0 - ln(2a) = -ln(2a)Putting it all together:f(x₁) + f(x₂) = (1/a - 1) - 2/a - ln(2a)Combine like terms:1/a - 1 - 2/a = (1/a - 2/a) - 1 = (-1/a) - 1So,f(x₁) + f(x₂) = (-1/a - 1) - ln(2a) = - (1/a + 1) - ln(2a)So,f(x₁) + f(x₂) = - (1 + 1/a) - ln(2a)We need to show that this is less than -3.So, we have:- (1 + 1/a) - ln(2a) < -3Multiply both sides by -1 (remembering to reverse the inequality):(1 + 1/a) + ln(2a) > 3So, we need to show that:1 + 1/a + ln(2a) > 3Simplify:1/a + ln(2a) > 2Let me define a function h(a) = 1/a + ln(2a). We need to show that h(a) > 2 for 0 < a < 1/2.Wait, but actually, we need to show that h(a) > 2, which would imply that -h(a) < -2, but wait, let's re-examine.Wait, from above:We have f(x₁) + f(x₂) = - (1 + 1/a) - ln(2a)We need to show that this is less than -3, so:- (1 + 1/a) - ln(2a) < -3Multiply both sides by -1 (inequality reverses):(1 + 1/a) + ln(2a) > 3So, yes, we need to show that 1 + 1/a + ln(2a) > 3, which simplifies to 1/a + ln(2a) > 2.So, define h(a) = 1/a + ln(2a). We need to show that h(a) > 2 for 0 < a < 1/2.Let's analyze h(a).First, note that a is in (0, 1/2).Compute h(a):h(a) = 1/a + ln(2a)We can write ln(2a) as ln 2 + ln a.So,h(a) = 1/a + ln 2 + ln aSo,h(a) = 1/a + ln a + ln 2We need to show that h(a) > 2.Let's compute the derivative of h(a) to see its behavior.h'(a) = derivative of 1/a + derivative of ln a + derivative of ln 2Which is:h'(a) = -1/a² + 1/a + 0 = (-1/a²) + (1/a)Simplify:h'(a) = (-1 + a)/a²Set derivative equal to zero to find critical points:(-1 + a)/a² = 0Numerator must be zero:-1 + a = 0 => a = 1But our domain is 0 < a < 1/2, so a = 1 is outside the domain. Therefore, h(a) has no critical points in (0, 1/2). So, we can check the behavior of h(a) as a approaches 0+ and as a approaches 1/2.First, as a approaches 0+:1/a approaches +∞, ln a approaches -∞, but 1/a dominates, so h(a) approaches +∞.As a approaches 1/2:Compute h(1/2):h(1/2) = 1/(1/2) + ln(2*(1/2)) = 2 + ln(1) = 2 + 0 = 2So, h(a) approaches 2 as a approaches 1/2 from the left.Now, since h(a) is decreasing on (0, 1/2) because h'(a) = (-1 + a)/a². Let's check the sign of h'(a) in (0, 1/2):For 0 < a < 1/2, numerator is (-1 + a). Since a < 1/2, (-1 + a) < -1 + 1/2 = -1/2 < 0. Denominator a² is positive. So, h'(a) < 0 in (0, 1/2). Therefore, h(a) is decreasing on (0, 1/2).Since h(a) is decreasing on (0, 1/2) and approaches +∞ as a approaches 0+, and approaches 2 as a approaches 1/2, then for all a in (0, 1/2), h(a) > 2.Therefore, h(a) = 1/a + ln(2a) > 2 for all 0 < a < 1/2.Thus, going back:1/a + ln(2a) > 2Therefore,1 + 1/a + ln(2a) > 3Which implies,- (1 + 1/a) - ln(2a) < -3Hence,f(x₁) + f(x₂) < -3Which is what we needed to prove.So, in summary, for part (2), by expressing f(x₁) + f(x₂) in terms of a, and showing that the resulting expression is always less than -3 for 0 < a < 1/2, we have proven the required inequality.

💡First, I need to determine the total amount of money Mrs. Petersons paid. She used five 100 bills, so the total payment is 5 multiplied by 100, which equals 500.Next, she received 50 in change. To find out how much money she actually spent on the tumblers, I subtract the change from the total payment: 500 minus 50 equals 450.Each tumbler costs 45. To find out how many tumblers she bought, I divide the total amount spent by the cost per tumbler: 450 divided by 45 equals 10.Therefore, Mrs. Petersons bought 10 tumblers.

💡Okay, so I have this problem where I need to find the intersection of two sets, A and B. Set A is defined as all x such that the absolute value of x minus 3 is less than or equal to 1. Set B is defined as all x such that x squared minus 5x plus 4 is greater than or equal to 0. I need to find A intersect B, which means I need to find all the x values that are in both sets A and B.First, let me tackle set A. The definition is |x - 3| ≤ 1. I remember that absolute value inequalities can be rewritten without the absolute value by considering the distance from a point. So, |x - 3| ≤ 1 means that x is within 1 unit of 3 on the number line. That translates to x being between 3 - 1 and 3 + 1, which is 2 ≤ x ≤ 4. So, set A is the interval [2, 4].Now, moving on to set B. It's defined by the inequality x² - 5x + 4 ≥ 0. Hmm, quadratic inequalities can be a bit tricky, but I think I can handle this. First, I should factor the quadratic expression if possible. Let me see if x² - 5x + 4 factors nicely. Looking for two numbers that multiply to 4 and add up to -5. Hmm, -1 and -4? Yes, because (-1) * (-4) = 4 and (-1) + (-4) = -5. So, the quadratic factors as (x - 1)(x - 4) ≥ 0.Now, to solve the inequality (x - 1)(x - 4) ≥ 0, I need to find the intervals where the product is positive or zero. The critical points are x = 1 and x = 4, which divide the number line into three intervals: (-∞, 1), (1, 4), and (4, ∞). I'll test each interval to see the sign of the product.1. For x < 1, let's pick x = 0. Plugging in, (0 - 1)(0 - 4) = (-1)(-4) = 4, which is positive. So, the inequality holds in this interval.2. For 1 < x < 4, let's pick x = 2. (2 - 1)(2 - 4) = (1)(-2) = -2, which is negative. So, the inequality does not hold here.3. For x > 4, let's pick x = 5. (5 - 1)(5 - 4) = (4)(1) = 4, which is positive. So, the inequality holds in this interval.Also, at the critical points x = 1 and x = 4, the expression equals zero, which satisfies the inequality since it's "greater than or equal to zero." Therefore, set B is the union of two intervals: (-∞, 1] and [4, ∞).Now, to find A ∩ B, I need to find the overlap between set A, which is [2, 4], and set B, which is (-∞, 1] ∪ [4, ∞). Let's visualize this on the number line.Set A spans from 2 to 4, and set B spans from negative infinity to 1 and from 4 to positive infinity. The only point where these two sets overlap is at x = 4 because that's the endpoint of set A and the starting point of the upper interval of set B.Wait, let me double-check that. Is there any overlap between [2, 4] and (-∞, 1]? Well, [2, 4] starts at 2, and (-∞, 1] ends at 1, so they don't overlap there. The other part of set B is [4, ∞), which starts at 4, and set A ends at 4. So, the only common point is 4 itself.Therefore, the intersection A ∩ B is just the single point {4}.I think that's it. Let me recap:- Set A: [2, 4]- Set B: (-∞, 1] ∪ [4, ∞)- Intersection: {4}Yes, that makes sense. I don't think I missed anything here. The key was to correctly solve both inequalities and then find where they overlap. Since set A is a closed interval and set B has two separate intervals, the only common point is at the boundary of set A and the start of the upper interval of set B.

💡First, I need to determine the total number of lollipops James has by adding up all the different types: cherry, wintergreen, grape, and shrimp cocktail lollipops.Next, I will divide the total number of lollipops by the number of friends (14) to find out how many lollipops each friend can receive evenly. The remainder from this division will indicate how many lollipops James has left after distributing them as much as possible.Finally, I will present the remainder as the number of lollipops James retains after the distribution.

💡Okay, so I need to find the volume of this solid. Let me try to visualize it first. The base is a triangle with sides s, s, and s√2. Hmm, that seems familiar. Maybe it's a right triangle? Let me check using the Pythagorean theorem. If it's a right triangle, then the sum of the squares of the two shorter sides should equal the square of the longest side.So, let's compute:s² + s² = 2s²And the square of the longest side is (s√2)² = 2s²Oh, okay, so yes, it is a right triangle. That makes things a bit easier because I can use the formula for the area of a right triangle, which is (base * height)/2.Given that s = 2√2, let me plug that in.First, the sides are s, s, and s√2, so substituting s:s = 2√2s√2 = 2√2 * √2 = 2 * 2 = 4So the sides are 2√2, 2√2, and 4. That confirms it's a right triangle because (2√2)² + (2√2)² = 8 + 8 = 16, and 4² is also 16.Now, the area of the base triangle is (base * height)/2. Since it's a right triangle, the two legs are perpendicular, so they can serve as base and height.So, area A = (2√2 * 2√2)/2 = (8)/2 = 4Wait, that seems straightforward. So the area of the base is 4.Now, the solid extends vertically upward from two opposite vertices with height h = 3s. Let me make sure I understand this correctly. It says two opposite vertices extend vertically upward by height h. So, does that mean it's like a prism where two vertices are lifted up to form a three-dimensional shape?Wait, if it's a prism, then the volume would be the area of the base times the height. But in this case, it's not a prism because only two vertices are extended upward. Hmm, maybe it's a kind of pyramid or something else.Wait, let me think again. If two opposite vertices are extended upward, perhaps it's forming a kind of wedge or a frustum? Or maybe it's a pyramid with a triangular base but only two vertices extended.Actually, now that I think about it, if two opposite vertices are extended upward by height h, maybe it's similar to extruding those two vertices into a line segment, creating a three-dimensional figure. But I'm not entirely sure.Wait, perhaps it's a kind of prismatoid. A prismatoid is a polyhedron whose vertices all lie in two parallel planes. In this case, the base is in one plane, and the top vertices are in another plane parallel to the base. But only two vertices are extended, so maybe it's a kind of pyramid with a triangular base but only two apex points?Wait, that might not make sense. Maybe it's better to think of it as a kind of extrusion where only two vertices are lifted, creating a slant.Alternatively, perhaps it's a kind of pyramid with a triangular base, but instead of a single apex, there are two apexes connected by an edge. That would make it a kind of bipyramid, but I'm not sure.Wait, maybe I'm overcomplicating it. Let me try to think differently. If two opposite vertices are extended upward by height h, then perhaps the solid is formed by connecting those two lifted vertices with the original base, creating a kind of ridge.Alternatively, maybe it's a triangular prism where two vertices are lifted, but that might not be a standard prism.Wait, perhaps it's a kind of extrusion where only two vertices are moved upward, keeping the third vertex at the base. That would create a three-dimensional figure with a triangular base and two vertical edges.Wait, let me try to sketch this mentally. The base is a right triangle. Let's label the vertices A, B, and C, where A and B are the vertices with the right angle, and C is the opposite vertex. If two opposite vertices are extended upward, perhaps A and C are extended upward by height h, while B remains at the base.So, vertex A is lifted to A', and vertex C is lifted to C', both at height h above the base. Then, the solid would have the original base triangle ABC, and the top triangle A'C' connected somehow.Wait, but if only A and C are lifted, then the figure would have edges AA' and CC', and perhaps a line connecting A' and C'. But what about the third vertex, B? It remains at the base, so maybe the solid is a kind of pyramid with base ABC and apexes at A' and C', connected by edges.Alternatively, perhaps it's a kind of prism where two vertices are lifted, but I'm not sure.Wait, maybe it's better to think of it as a kind of extrusion where two vertices are lifted, creating a slant. So, the solid would have the original base, and then two vertical edges from A and C, and then a line connecting A' and C' at the top. The other edges would be slant edges from B to A' and B to C'.Wait, that might form a kind of tetrahedron, but with two apexes. Hmm, maybe it's a kind of double pyramid.Alternatively, perhaps it's a kind of wedge. Let me think.Wait, maybe it's a kind of pyramid with a triangular base, but instead of a single apex, there are two apexes connected by an edge. So, the base is triangle ABC, and the top is a line segment A'C', with A' above A and C' above C, both at height h.In that case, the solid would have five faces: the base ABC, the two vertical faces AA'B and CC'B, and the two slant faces A'BC' and maybe A'CC' or something else.Wait, I'm getting confused. Maybe I should try to calculate the volume using coordinates.Let me assign coordinates to the vertices. Let me place the right triangle in the xy-plane with vertex A at (0,0,0), vertex B at (s, 0, 0), and vertex C at (0, s, 0). Wait, but given the sides are s, s, and s√2, that would make sense because the distance from A to B is s, from A to C is s, and from B to C is s√2.So, with s = 2√2, the coordinates would be:A: (0, 0, 0)B: (2√2, 0, 0)C: (0, 2√2, 0)Now, two opposite vertices are extended vertically upward by height h = 3s = 3*(2√2) = 6√2.Which two opposite vertices? In a triangle, opposite vertices would be connected by an edge. So, in triangle ABC, the vertices are A, B, and C. The opposite of A is BC, but vertices are points, so maybe the opposite vertices are A and C, or A and B, or B and C.Wait, in a triangle, each vertex is connected to the other two, so maybe "opposite" here refers to the two vertices not connected by an edge? But in a triangle, every pair of vertices is connected by an edge, so that doesn't make sense.Wait, maybe "opposite" here refers to the vertices that are not adjacent. But in a triangle, every vertex is adjacent to the other two, so that can't be.Wait, perhaps "opposite" refers to the vertices that are not the right angle. In a right triangle, the right angle is at A, so the other two vertices B and C are the ones opposite the right angle. So, maybe those two are extended upward.Wait, but the problem says "two opposite vertices of the triangle extend vertically upward". So, perhaps it's two vertices that are opposite each other in some sense. In a triangle, it's a bit ambiguous, but perhaps it's the two vertices that are not the right angle vertex.So, in my coordinate system, A is the right angle, so B and C are the other two vertices. So, maybe B and C are extended upward.Wait, but the problem says "two opposite vertices", so maybe it's A and C, or A and B. Hmm.Wait, perhaps it's better to assume that the two vertices that are not connected by the hypotenuse are extended. So, in the right triangle, the hypotenuse is BC, so the other two vertices are A and... Wait, no, A is connected to both B and C.Wait, maybe it's better to think that in the triangle, the two vertices that are endpoints of the hypotenuse are extended. So, if the hypotenuse is BC, then B and C are extended upward.Wait, but in that case, both B and C are extended upward, so their new positions would be B' and C' at (2√2, 0, h) and (0, 2√2, h).Alternatively, if A and C are extended, then A' would be at (0,0,h) and C' at (0,2√2,h). Similarly, if A and B are extended, A' at (0,0,h) and B' at (2√2,0,h).Wait, the problem says "two opposite vertices", so perhaps it's the two vertices that are not adjacent, but in a triangle, all vertices are adjacent. So, maybe it's the two vertices that are not the right angle. So, in my coordinate system, A is the right angle, so B and C are extended.So, B is at (2√2, 0, 0), and C is at (0, 2√2, 0). So, extending them upward by h = 6√2, their new positions would be B' at (2√2, 0, 6√2) and C' at (0, 2√2, 6√2).Now, the solid would have the original base triangle ABC, and the top triangle A'B'C', but wait, only B and C are extended, so A remains at (0,0,0). So, the solid would have vertices A, B, C, B', C'.Wait, but then how is the solid formed? It would have the base ABC, and then two vertical edges from B to B' and from C to C'. Then, the top would be a line segment from B' to C', and the sides would be triangles ABB', ACC', and the quadrilateral BB'C'C.Wait, but that would make the solid a kind of prismatoid with vertices A, B, C, B', C'. But since only two vertices are lifted, it's not a full prism.Alternatively, maybe it's a kind of pyramid with a triangular base and two apexes. But pyramids typically have a single apex.Wait, perhaps it's better to think of it as a kind of extrusion where only two vertices are moved, creating a slant. So, the volume can be calculated by integrating or using some geometric formulas.Alternatively, maybe it's a kind of wedge. Let me think about the shape.Alternatively, perhaps it's a kind of pyramid with a triangular base and two apexes connected by an edge. So, the base is triangle ABC, and the top is a line segment B'C', with B' above B and C' above C.In that case, the solid would have five faces: the base ABC, the two vertical faces ABB' and ACC', and the two slant faces BB'C' and CC'A.Wait, but I'm not sure. Maybe it's better to calculate the volume using coordinates.So, let's assign coordinates as follows:A: (0, 0, 0)B: (2√2, 0, 0)C: (0, 2√2, 0)B': (2√2, 0, 6√2)C': (0, 2√2, 6√2)Now, the solid is formed by connecting these points. So, the solid has vertices A, B, C, B', C'.To find the volume, perhaps we can use the formula for the volume of a polyhedron given its vertices. Alternatively, we can divide the solid into simpler parts whose volumes we can calculate and then sum them up.Alternatively, since it's a kind of prismatoid, we can use the formula for the volume of a prismatoid, which is:V = (h/6) * (A1 + A2 + 4M)where A1 and A2 are the areas of the two parallel faces, and M is the area of the midsection.But in this case, I'm not sure if the two bases are parallel. Wait, the base is triangle ABC, and the top is a line segment B'C'. So, the top is not a face but a line segment. So, maybe the formula doesn't apply.Alternatively, perhaps we can think of the solid as a combination of two pyramids.Wait, let me think. If we consider the solid as having the base ABC and two apexes B' and C', then perhaps it's a kind of double pyramid. But I'm not sure.Alternatively, maybe it's better to use the divergence theorem or integrate, but that might be overkill.Wait, perhaps it's better to think of the solid as a kind of extrusion where two vertices are lifted, creating a slant. So, the volume can be calculated as the area of the base times the average height.Wait, but the height is not uniform because only two vertices are lifted. So, maybe the average height is h/2, but I'm not sure.Wait, let me think differently. If I can find the height function over the base triangle, then integrate it to find the volume.But that might be complicated. Alternatively, perhaps we can use the formula for the volume of a solid with a triangular base and linearly varying height.Wait, in this case, the height at points B and C is h, and at point A, it's 0. So, the height varies linearly across the base.Wait, that might be the case. So, the height at any point (x, y) in the base triangle can be expressed as a linear function.Given that, the volume can be calculated as the double integral over the base of the height function.So, let's define the height function h(x, y). At point A (0,0), h=0. At point B (2√2, 0), h=6√2. At point C (0, 2√2), h=6√2.So, the height function is linear, meaning it can be expressed as h(x, y) = ax + by + c.We can find a, b, c using the given points.At A (0,0): h=0 => 0 = a*0 + b*0 + c => c=0.At B (2√2, 0): h=6√2 => 6√2 = a*(2√2) + b*0 => 6√2 = 2√2 a => a = 3.At C (0, 2√2): h=6√2 => 6√2 = a*0 + b*(2√2) => 6√2 = 2√2 b => b=3.So, the height function is h(x, y) = 3x + 3y.Now, the volume is the double integral over the base triangle of h(x, y) dA.So, V = ∫∫ h(x, y) dA over the base triangle.Since the base is a right triangle with vertices at (0,0), (2√2, 0), and (0, 2√2), we can set up the integral in the xy-plane.Let me set up the limits. x goes from 0 to 2√2, and for each x, y goes from 0 to (2√2 - x), because the hypotenuse is the line x + y = 2√2.So, V = ∫ (x=0 to 2√2) ∫ (y=0 to 2√2 - x) (3x + 3y) dy dxLet me compute the inner integral first:∫ (y=0 to 2√2 - x) (3x + 3y) dy= ∫ (3x + 3y) dy from 0 to 2√2 - x= 3x * y + (3/2)y² evaluated from 0 to 2√2 - x= 3x*(2√2 - x) + (3/2)*(2√2 - x)²Now, let's compute this:First term: 3x*(2√2 - x) = 6√2 x - 3x²Second term: (3/2)*(2√2 - x)²Let me expand (2√2 - x)²:= (2√2)² - 2*(2√2)*x + x²= 8 - 4√2 x + x²So, the second term becomes (3/2)*(8 - 4√2 x + x²) = 12 - 6√2 x + (3/2)x²Now, combining both terms:6√2 x - 3x² + 12 - 6√2 x + (3/2)x²Simplify:6√2 x - 6√2 x = 0-3x² + (3/2)x² = (-3 + 1.5)x² = (-1.5)x² = (-3/2)x²And the constant term is 12.So, the inner integral simplifies to 12 - (3/2)x²Now, the volume integral becomes:V = ∫ (x=0 to 2√2) [12 - (3/2)x²] dxLet's compute this integral:= ∫12 dx - (3/2)∫x² dx from 0 to 2√2= 12x - (3/2)*(x³/3) evaluated from 0 to 2√2Simplify:= 12x - (1/2)x³ evaluated from 0 to 2√2Now, plug in x = 2√2:= 12*(2√2) - (1/2)*(2√2)³Compute each term:12*(2√2) = 24√2(2√2)³ = 8*(√2)³ = 8*(2√2) = 16√2So, (1/2)*(16√2) = 8√2Therefore, the expression becomes:24√2 - 8√2 = 16√2Now, plug in x=0, which gives 0 - 0 = 0So, the volume V = 16√2 - 0 = 16√2Wait, but earlier I thought the area of the base was 4, and if it were a prism, the volume would be 4 * 6√2 = 24√2. But here, using the integral, I got 16√2. So, there's a discrepancy.Wait, maybe I made a mistake in setting up the height function. Let me double-check.I assumed that the height at A is 0, and at B and C it's 6√2. So, h(x, y) = 3x + 3y.But when I integrated this over the base, I got 16√2, which is less than the prism volume of 24√2.Wait, perhaps the solid is not a linear extrusion but something else. Maybe the height is not varying linearly across the base.Wait, let me think again. If only two vertices are extended upward, then the height at those two points is h, and at the third vertex, it's 0. So, the height function should be linear across the base, connecting these three points.Wait, but in my integral, I considered h(x, y) = 3x + 3y, which gives h=6√2 at both B and C, and h=0 at A. That seems correct.But the integral gave me 16√2, while the prism volume is 24√2. So, why the difference?Wait, maybe I made a mistake in the integral calculation. Let me check the steps again.First, the inner integral:∫ (y=0 to 2√2 - x) (3x + 3y) dy= 3x*(2√2 - x) + (3/2)*(2√2 - x)²= 6√2 x - 3x² + (3/2)*(8 - 4√2 x + x²)= 6√2 x - 3x² + 12 - 6√2 x + (3/2)x²= (6√2 x - 6√2 x) + (-3x² + 1.5x²) + 12= 0 - 1.5x² + 12= 12 - 1.5x²Then, the outer integral:∫ (x=0 to 2√2) (12 - 1.5x²) dx= 12x - 0.5x³ evaluated from 0 to 2√2At x=2√2:12*(2√2) = 24√20.5*(2√2)^3 = 0.5*(16√2) = 8√2So, 24√2 - 8√2 = 16√2Hmm, that seems correct. So, the volume is 16√2.But earlier, I thought it might be 24√2 if it were a prism. So, why the difference?Wait, perhaps because in a prism, all points are lifted uniformly, but in this case, only two vertices are lifted, so the height varies across the base, leading to a smaller volume.Wait, but let me think again. If two vertices are lifted, and the third remains, then the solid is a kind of pyramid with a triangular base and two apexes. But pyramids typically have a single apex, so this might be a kind of double pyramid or a bifrustum.Alternatively, perhaps it's a kind of extrusion where the height varies linearly from 0 at A to h at B and C.Wait, in that case, the volume would indeed be the integral of the height function over the base, which we calculated as 16√2.But let me think if there's another way to calculate this volume without integration.Wait, perhaps we can think of the solid as a combination of two tetrahedrons.Let me consider the solid as two tetrahedrons: one with base ABC and apex B', and another with base ABC and apex C'.But wait, no, because B' and C' are both lifted from B and C, so the solid is actually a kind of bipyramid.Wait, a bipyramid is formed by joining two pyramids at their base. But in this case, the base is a triangle, and the two apexes are B' and C'.Wait, but in a bipyramid, the apexes are on opposite sides of the base, but here, B' and C' are both above the base in the same direction.Wait, maybe it's better to think of it as a kind of wedge.Alternatively, perhaps we can use the formula for the volume of a tetrahedron.Wait, a tetrahedron has four triangular faces. In this case, our solid has five faces: the base ABC, the two vertical faces ABB' and ACC', and the two slant faces BB'C' and CC'A.Wait, actually, the solid has five vertices: A, B, C, B', C'.Wait, perhaps it's a kind of pyramid with a quadrilateral base. But no, the base is a triangle.Wait, maybe it's a kind of prism with a triangular base but only two vertices lifted, making it a kind of irregular prism.Alternatively, perhaps it's better to use the shoelace formula or some other method.Wait, another approach: since the solid is formed by lifting two vertices of the base triangle, the volume can be calculated as the area of the base times the average height.But what is the average height?In this case, the height at A is 0, and at B and C, it's 6√2. So, the average height might be (0 + 6√2 + 6√2)/3 = (12√2)/3 = 4√2.Then, volume would be base area * average height = 4 * 4√2 = 16√2, which matches the integral result.So, that seems to confirm the volume is 16√2.Wait, but earlier I thought it might be 24√2 if it were a prism, but in this case, since only two vertices are lifted, the average height is less, leading to a smaller volume.So, perhaps the correct volume is 16√2.But wait, let me think again. If I consider the solid as a kind of extrusion where the height varies linearly from 0 at A to 6√2 at B and C, then the average height is indeed 4√2, so the volume is 4 * 4√2 = 16√2.Alternatively, if it were a prism, all points would be lifted uniformly, so the height would be 6√2 everywhere, leading to volume 4 * 6√2 = 24√2.But in this case, since only two vertices are lifted, the height varies, so the volume is less.Therefore, the volume of the solid is 16√2.Wait, but earlier I got 16√2 using integration, and also by considering the average height. So, that seems consistent.But let me check once more.Alternatively, perhaps the solid is a kind of pyramid with a triangular base and two apexes. Wait, but pyramids have a single apex. So, maybe it's a kind of double pyramid, but I'm not sure.Alternatively, perhaps it's a kind of frustum, but a frustum is a portion of a solid, usually a cone or pyramid, between two parallel planes. But in this case, the top is not a similar triangle but a line segment.Wait, perhaps it's better to think of it as a kind of extrusion where the height varies linearly from 0 at A to 6√2 at B and C.In that case, the volume would indeed be the integral of the height function over the base, which we calculated as 16√2.So, I think the correct volume is 16√2.Wait, but earlier I thought the answer was 24√2, but that was under the assumption that it's a prism, which it's not. Since only two vertices are lifted, the volume is less.Therefore, the volume of the solid is 16√2.