💡Alright, so I have this problem about an ellipse in the Cartesian coordinate system. Let me try to break it down step by step. First, part (I) asks for the standard equation of the ellipse C. The given equation is (frac{x^2}{a^2} + frac{y^2}{b^2} = 1), where (a > b > 0). The left focus is F, the eccentricity is (frac{sqrt{2}}{2}), and the length of the chord passing through F and perpendicular to the major axis is (sqrt{2}).Okay, so I remember that for an ellipse, the eccentricity (e) is given by (e = frac{c}{a}), where (c) is the distance from the center to each focus. Since the eccentricity is (frac{sqrt{2}}{2}), that means (c = a cdot frac{sqrt{2}}{2}).Also, the relationship between (a), (b), and (c) is (c^2 = a^2 - b^2). So, substituting (c) from above, we have:[left(a cdot frac{sqrt{2}}{2}right)^2 = a^2 - b^2][frac{2a^2}{4} = a^2 - b^2][frac{a^2}{2} = a^2 - b^2][b^2 = a^2 - frac{a^2}{2} = frac{a^2}{2}]So, (b = frac{a}{sqrt{2}}).Now, the chord passing through the focus F and perpendicular to the major axis is called the latus rectum. The length of the latus rectum in an ellipse is given by (frac{2b^2}{a}). According to the problem, this length is (sqrt{2}). So,[frac{2b^2}{a} = sqrt{2}]But we already know that (b^2 = frac{a^2}{2}), so substituting that in:[frac{2 cdot frac{a^2}{2}}{a} = sqrt{2}][frac{a^2}{a} = sqrt{2}][a = sqrt{2}]So, (a = sqrt{2}), and since (b^2 = frac{a^2}{2}), substituting (a):[b^2 = frac{(sqrt{2})^2}{2} = frac{2}{2} = 1]Thus, (b = 1).Therefore, the standard equation of the ellipse is:[frac{x^2}{2} + y^2 = 1]Alright, that seems solid. Let me just double-check my steps. Eccentricity given, found (c) in terms of (a), used the relationship (c^2 = a^2 - b^2) to find (b) in terms of (a), then used the latus rectum length to solve for (a). Seems good.Moving on to part (II). Points A and B are the left and right vertices of the ellipse, respectively. So, A is at ((-a, 0)) which is ((- sqrt{2}, 0)), and B is at ((sqrt{2}, 0)). A line passes through point P(-2, 0) and intersects the ellipse at two distinct points M and N.Part (i) asks to prove that (angle AFM = angle BFN). Hmm, okay. So, F is the left focus. From part (I), we know that (c = a cdot frac{sqrt{2}}{2} = sqrt{2} cdot frac{sqrt{2}}{2} = 1). So, the foci are at ((-c, 0)) and ((c, 0)), which is ((-1, 0)) and ((1, 0)). So, F is at (-1, 0).So, points A(-√2, 0), F(-1, 0), B(√2, 0). The line passes through P(-2, 0) and intersects the ellipse at M and N. We need to show that the angles AFM and BFN are equal.Hmm, maybe I can use some properties of ellipses or coordinate geometry here. Since the line passes through P(-2, 0), which is outside the ellipse, it intersects the ellipse at two points M and N. So, maybe I can parametrize the line and find the coordinates of M and N, then compute the angles.Alternatively, perhaps there's a symmetry or reflection property that can be used. Since F is a focus, maybe the reflection property of ellipses can help. The reflection property states that the angle between the tangent at any point on the ellipse and the line from that point to one focus is equal to the angle between the tangent and the line to the other focus. But here, we're dealing with chords, not tangents.Wait, but maybe if I consider the line through P intersecting the ellipse at M and N, then perhaps the angles at F from A to M and from B to N are equal. Maybe by showing that the slopes of FM and FN have some relationship, or that the triangles AFM and BFN are similar or something.Let me try to parametrize the line. Let's assume the line passing through P(-2, 0) has a slope m. So, the equation of the line is (y = m(x + 2)). This line intersects the ellipse (frac{x^2}{2} + y^2 = 1). Substituting y into the ellipse equation:[frac{x^2}{2} + (m(x + 2))^2 = 1][frac{x^2}{2} + m^2(x^2 + 4x + 4) = 1]Multiply through by 2 to eliminate the fraction:[x^2 + 2m^2(x^2 + 4x + 4) = 2][x^2 + 2m^2x^2 + 8m^2x + 8m^2 = 2]Combine like terms:[(1 + 2m^2)x^2 + 8m^2x + (8m^2 - 2) = 0]This is a quadratic in x. Let me denote this as:[Ax^2 + Bx + C = 0]where (A = 1 + 2m^2), (B = 8m^2), and (C = 8m^2 - 2).The solutions will give the x-coordinates of M and N. Let me denote them as (x_1) and (x_2). Then, from Vieta's formulas:[x_1 + x_2 = -frac{B}{A} = -frac{8m^2}{1 + 2m^2}][x_1 x_2 = frac{C}{A} = frac{8m^2 - 2}{1 + 2m^2}]Now, the corresponding y-coordinates are (y_1 = m(x_1 + 2)) and (y_2 = m(x_2 + 2)).Now, to find angles AFM and BFN. Let me think about how to compute these angles.Point A is (-√2, 0), F is (-1, 0), and M is (x1, y1). Similarly, point B is (√2, 0), F is (-1, 0), and N is (x2, y2).So, angle AFM is the angle at F between points A and M. Similarly, angle BFN is the angle at F between points B and N.To compute these angles, perhaps I can compute the slopes of lines FA and FM, and similarly for FB and FN, then find the angles between these lines.But wait, FA is the line from F(-1, 0) to A(-√2, 0). Since both are on the x-axis, FA is just a horizontal line to the left. Similarly, FB is the line from F(-1, 0) to B(√2, 0), which is a horizontal line to the right.So, angle AFM is the angle between FA (which is along the negative x-axis) and FM (which goes from F(-1, 0) to M(x1, y1)). Similarly, angle BFN is the angle between FB (along the positive x-axis) and FN (from F(-1, 0) to N(x2, y2)).So, to find these angles, I can compute the slopes of FM and FN, and then find the angles they make with FA and FB, respectively.Alternatively, since FA is along the negative x-axis, the angle AFM is the angle between the negative x-axis and the line FM. Similarly, angle BFN is the angle between the positive x-axis and FN.But since we're dealing with angles, perhaps using vectors would be better. The angle between two vectors can be found using the dot product.Let me denote vector FA as from F to A: A - F = (-√2 - (-1), 0 - 0) = (-√2 + 1, 0). Similarly, vector FM is M - F = (x1 - (-1), y1 - 0) = (x1 + 1, y1).Similarly, vector FB is B - F = (√2 - (-1), 0 - 0) = (√2 + 1, 0). Vector FN is N - F = (x2 + 1, y2).The angle between FA and FM is the angle between vectors FA and FM. Similarly, the angle between FB and FN is the angle between vectors FB and FN.But since FA is (-√2 + 1, 0) and FB is (√2 + 1, 0), which are both along the x-axis, the angles AFM and BFN can be found by the angles that FM and FN make with the x-axis.Wait, but FA is along the negative x-axis, so the angle AFM is measured from the negative x-axis to FM. Similarly, FB is along the positive x-axis, so angle BFN is measured from the positive x-axis to FN.But since the line passes through P(-2, 0), which is to the left of F(-1, 0), the points M and N are likely on either side of F? Or maybe both on one side? Hmm, not sure.Wait, actually, since the line passes through P(-2, 0), which is outside the ellipse, it will intersect the ellipse at two points M and N. Depending on the slope, these points can be on either side of F.But regardless, to find the angles AFM and BFN, perhaps I can compute the slopes of FM and FN and then find the angles they make with FA and FB.Alternatively, since FA is along the negative x-axis, the slope of FA is 0 (it's horizontal). The slope of FM is (y1 - 0)/(x1 - (-1)) = y1/(x1 + 1). Similarly, the slope of FN is y2/(x2 + 1).The angle between FA and FM is the angle whose tangent is equal to the slope of FM. Similarly, the angle between FB and FN is the angle whose tangent is equal to the slope of FN.But since FA is along the negative x-axis, the angle AFM is actually the angle between the negative x-axis and FM, which would be 180 degrees minus the angle that FM makes with the positive x-axis. Similarly, angle BFN is the angle between the positive x-axis and FN.Wait, this is getting a bit confusing. Maybe I should compute the actual angles using arctangent.Let me denote θ1 as angle AFM and θ2 as angle BFN.Then,[tan(theta1) = left|frac{y1}{x1 + 1}right|][tan(theta2) = left|frac{y2}{x2 + 1}right|]We need to show that θ1 = θ2, which would mean that (tan(theta1) = tan(theta2)), so (left|frac{y1}{x1 + 1}right| = left|frac{y2}{x2 + 1}right|).But since the line passes through P(-2, 0), and the ellipse is symmetric about the x-axis, perhaps the points M and N are symmetric in some way.Wait, but the line has a slope m, so unless m is zero, the points M and N won't be symmetric about the x-axis. Hmm.Alternatively, maybe the product of the slopes of FM and FN is -1, meaning they are perpendicular, but that doesn't necessarily make the angles equal.Wait, let me think differently. Maybe the triangles AFM and BFN are similar.For triangles AFM and BFN to be similar, their corresponding angles must be equal. If we can show that the sides are proportional, that might help.But I'm not sure. Maybe another approach is needed.Wait, another idea: since the line passes through P(-2, 0), which is outside the ellipse, and F is at (-1, 0), maybe there's a harmonic division or something related to projective geometry. But that might be too advanced.Alternatively, perhaps using parametric equations for the ellipse.The ellipse equation is (frac{x^2}{2} + y^2 = 1). So, parametric equations can be written as:[x = sqrt{2} cos theta][y = sin theta]So, any point on the ellipse can be represented as ((sqrt{2} cos theta, sin theta)). The line passing through P(-2, 0) can be parametrized as (y = m(x + 2)). So, substituting into the ellipse equation:[frac{x^2}{2} + m^2(x + 2)^2 = 1]Which is the same equation I had before. So, solving for x gives the points M and N.But maybe instead of using slope m, I can use parametric coordinates.Alternatively, maybe using the concept of pole and polar. The polar line of point P with respect to the ellipse might have some relation.But perhaps that's complicating things.Wait, another idea: since we need to show that angles AFM and BFN are equal, maybe the lines FM and FN are symmetric with respect to the x-axis or something.But since the line passes through P(-2, 0), which is on the x-axis, perhaps the points M and N are such that their y-coordinates are negatives of each other, but that would only be if the line is horizontal, which it's not necessarily.Wait, unless the line is not horizontal, but still, the y-coordinates might not be negatives.Alternatively, maybe the product of the slopes of FM and FN is 1, meaning they are reciprocals, but that would mean the angles are complementary, not necessarily equal.Wait, perhaps I can compute the slopes of FM and FN and see if they satisfy some relationship.From earlier, we have:Slope of FM: (k1 = frac{y1}{x1 + 1})Slope of FN: (k2 = frac{y2}{x2 + 1})We need to show that the angles between FA and FM, and between FB and FN are equal. Since FA is along the negative x-axis and FB is along the positive x-axis, the angles θ1 and θ2 can be related to the slopes k1 and k2.But perhaps instead of dealing with angles, I can consider the tangent of the angles and show that they are equal in magnitude.So, (tan(theta1) = |k1|) and (tan(theta2) = |k2|). So, if I can show that |k1| = |k2|, then θ1 = θ2.So, let's compute k1 and k2.From earlier, we have:(y1 = m(x1 + 2))So,(k1 = frac{m(x1 + 2)}{x1 + 1})Similarly,(k2 = frac{m(x2 + 2)}{x2 + 1})So, we need to show that:[left|frac{m(x1 + 2)}{x1 + 1}right| = left|frac{m(x2 + 2)}{x2 + 1}right|]Since m is non-zero (otherwise, the line would be horizontal and might not intersect the ellipse at two distinct points unless m=0, but let's assume m ≠ 0 for now), we can cancel m:[left|frac{x1 + 2}{x1 + 1}right| = left|frac{x2 + 2}{x2 + 1}right|]So, we need to show that:[frac{x1 + 2}{x1 + 1} = pm frac{x2 + 2}{x2 + 1}]But since the line intersects the ellipse at M and N, which are two distinct points, and given the symmetry, perhaps the expressions are equal in magnitude but opposite in sign.Wait, let's compute (frac{x1 + 2}{x1 + 1}) and (frac{x2 + 2}{x2 + 1}).Let me denote (t1 = frac{x1 + 2}{x1 + 1}) and (t2 = frac{x2 + 2}{x2 + 1}).We need to show that |t1| = |t2|.Alternatively, maybe t1 = -t2, which would make their magnitudes equal.Let me compute t1 + t2 and t1 * t2.First, t1 + t2:[t1 + t2 = frac{x1 + 2}{x1 + 1} + frac{x2 + 2}{x2 + 1}]Let me combine these fractions:[= frac{(x1 + 2)(x2 + 1) + (x2 + 2)(x1 + 1)}{(x1 + 1)(x2 + 1)}]Expanding the numerator:[= frac{x1x2 + x1 + 2x2 + 2 + x1x2 + x2 + 2x1 + 2}{(x1 + 1)(x2 + 1)}][= frac{2x1x2 + 3x1 + 3x2 + 4}{(x1 + 1)(x2 + 1)}]Now, from earlier, we have:(x1 + x2 = -frac{8m^2}{1 + 2m^2})and(x1 x2 = frac{8m^2 - 2}{1 + 2m^2})So, let's substitute these into the numerator:Numerator:[2x1x2 + 3(x1 + x2) + 4 = 2 cdot frac{8m^2 - 2}{1 + 2m^2} + 3 cdot left(-frac{8m^2}{1 + 2m^2}right) + 4][= frac{16m^2 - 4}{1 + 2m^2} - frac{24m^2}{1 + 2m^2} + 4][= frac{16m^2 - 4 - 24m^2}{1 + 2m^2} + 4][= frac{-8m^2 - 4}{1 + 2m^2} + 4][= frac{-8m^2 - 4 + 4(1 + 2m^2)}{1 + 2m^2}][= frac{-8m^2 - 4 + 4 + 8m^2}{1 + 2m^2}][= frac{0}{1 + 2m^2} = 0]So, the numerator is zero, which means t1 + t2 = 0.Therefore, t1 = -t2.So, (frac{x1 + 2}{x1 + 1} = -frac{x2 + 2}{x2 + 1})Therefore, (left|frac{x1 + 2}{x1 + 1}right| = left|frac{x2 + 2}{x2 + 1}right|)Which implies that |k1| = |k2|, so (tan(theta1) = tan(theta2)), meaning θ1 = θ2 or θ1 = π - θ2. But since both angles are between 0 and π, and given the configuration, they should be equal.Therefore, (angle AFM = angle BFN).Okay, that seems to work. So, part (i) is proved.Now, part (ii) asks to find the maximum area of triangle MNF.So, points M and N are on the ellipse, and F is (-1, 0). We need to find the maximum area of triangle MNF as the line through P(-2, 0) varies.The area of triangle given three points can be found using the determinant formula:[text{Area} = frac{1}{2} |(x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2))|]Here, points are M(x1, y1), N(x2, y2), and F(-1, 0). So,[text{Area} = frac{1}{2} |x1(y2 - 0) + x2(0 - y1) + (-1)(y1 - y2)|][= frac{1}{2} |x1 y2 - x2 y1 - y1 + y2|][= frac{1}{2} |(x1 y2 - x2 y1) + (y2 - y1)|]Hmm, that seems a bit complicated. Maybe another approach is better.Alternatively, since F is at (-1, 0), and M and N are points on the ellipse, the area can be expressed as half the base times height. But since the base is MN and the height is the distance from F to the line MN, which is fixed as the line passes through P(-2, 0). Wait, no, the line MN passes through P, but F is fixed.Wait, actually, the area of triangle MNF can be found using the formula:[text{Area} = frac{1}{2} | vec{FM} times vec{FN} |]Where (vec{FM}) and (vec{FN}) are vectors from F to M and N, respectively. The cross product magnitude gives the area of the parallelogram, so half of that is the area of the triangle.So, let's compute vectors FM and FN.FM = (x1 + 1, y1 - 0) = (x1 + 1, y1)FN = (x2 + 1, y2 - 0) = (x2 + 1, y2)The cross product in 2D is scalar and given by:[vec{FM} times vec{FN} = (x1 + 1)(y2) - (x2 + 1)(y1)]So, the area is:[frac{1}{2} |(x1 + 1)y2 - (x2 + 1)y1|]Which is the same as the determinant approach.So, let's express this in terms of m.We have y1 = m(x1 + 2) and y2 = m(x2 + 2).So,[text{Area} = frac{1}{2} |(x1 + 1)m(x2 + 2) - (x2 + 1)m(x1 + 2)|][= frac{m}{2} |(x1 + 1)(x2 + 2) - (x2 + 1)(x1 + 2)|]Let me expand the terms inside the absolute value:First term: (x1 + 1)(x2 + 2) = x1x2 + 2x1 + x2 + 2Second term: (x2 + 1)(x1 + 2) = x1x2 + 2x2 + x1 + 2Subtracting the second term from the first:[(x1x2 + 2x1 + x2 + 2) - (x1x2 + 2x2 + x1 + 2) = (2x1 + x2) - (2x2 + x1) = x1 - x2]So, the area becomes:[frac{m}{2} |x1 - x2|]But since area is positive, we can drop the absolute value:[text{Area} = frac{|m|}{2} |x1 - x2|]Now, |x1 - x2| can be found from the quadratic equation we had earlier. For a quadratic equation (Ax^2 + Bx + C = 0), the difference of roots is (sqrt{(x1 + x2)^2 - 4x1x2}).So,[|x1 - x2| = sqrt{(x1 + x2)^2 - 4x1x2}]From earlier, we have:(x1 + x2 = -frac{8m^2}{1 + 2m^2})and(x1 x2 = frac{8m^2 - 2}{1 + 2m^2})So,[|x1 - x2| = sqrt{left(-frac{8m^2}{1 + 2m^2}right)^2 - 4 cdot frac{8m^2 - 2}{1 + 2m^2}}][= sqrt{frac{64m^4}{(1 + 2m^2)^2} - frac{32m^2 - 8}{1 + 2m^2}}][= sqrt{frac{64m^4 - (32m^2 - 8)(1 + 2m^2)}{(1 + 2m^2)^2}}]Let me compute the numerator:[64m^4 - (32m^2 - 8)(1 + 2m^2)][= 64m^4 - [32m^2(1) + 32m^2(2m^2) - 8(1) - 8(2m^2)]][= 64m^4 - [32m^2 + 64m^4 - 8 - 16m^2]][= 64m^4 - [ (32m^2 - 16m^2) + 64m^4 - 8 ]][= 64m^4 - [16m^2 + 64m^4 - 8]][= 64m^4 - 16m^2 - 64m^4 + 8][= -16m^2 + 8][= 8 - 16m^2]So, the numerator is (8 - 16m^2), and the denominator is ((1 + 2m^2)^2). Therefore,[|x1 - x2| = sqrt{frac{8 - 16m^2}{(1 + 2m^2)^2}} = frac{sqrt{8 - 16m^2}}{1 + 2m^2}]But for the square root to be real, the numerator must be non-negative:[8 - 16m^2 geq 0 implies 16m^2 leq 8 implies m^2 leq frac{1}{2} implies |m| leq frac{sqrt{2}}{2}]So, the area becomes:[text{Area} = frac{|m|}{2} cdot frac{sqrt{8 - 16m^2}}{1 + 2m^2}][= frac{|m| sqrt{8 - 16m^2}}{2(1 + 2m^2)}][= frac{|m| sqrt{8(1 - 2m^2)}}{2(1 + 2m^2)}][= frac{|m| cdot 2sqrt{2} sqrt{1 - 2m^2}}{2(1 + 2m^2)}][= frac{|m| sqrt{2} sqrt{1 - 2m^2}}{1 + 2m^2}]Since area is positive, we can drop the absolute value on m:[text{Area} = frac{m sqrt{2} sqrt{1 - 2m^2}}{1 + 2m^2}]But m can be positive or negative, but since we're dealing with area, which is positive, we can consider m > 0 without loss of generality.So, let me set (m > 0), then:[text{Area} = frac{m sqrt{2} sqrt{1 - 2m^2}}{1 + 2m^2}]Now, to find the maximum area, we can consider this as a function of m:[A(m) = frac{m sqrt{2} sqrt{1 - 2m^2}}{1 + 2m^2}]We need to maximize A(m) with respect to m, where (0 < m leq frac{sqrt{2}}{2}).Let me set (t = m^2), so (0 < t leq frac{1}{2}).Then,[A(t) = frac{sqrt{t} sqrt{2} sqrt{1 - 2t}}{1 + 2t}][= sqrt{2} cdot frac{sqrt{t(1 - 2t)}}{1 + 2t}]Let me square the area to make it easier to maximize:[A^2(t) = 2 cdot frac{t(1 - 2t)}{(1 + 2t)^2}]Let me denote (f(t) = frac{t(1 - 2t)}{(1 + 2t)^2}). We need to maximize f(t).So,[f(t) = frac{t - 2t^2}{(1 + 2t)^2}]To find the maximum, take the derivative of f(t) with respect to t and set it to zero.First, compute f'(t):Using the quotient rule:[f'(t) = frac{(1 - 4t)(1 + 2t)^2 - (t - 2t^2)(2)(1 + 2t)(2)}{(1 + 2t)^4}]Wait, that seems complicated. Let me simplify step by step.Let me denote numerator as N = t - 2t^2 and denominator as D = (1 + 2t)^2.Then,f(t) = N / Df'(t) = (N' D - N D') / D^2Compute N':N = t - 2t^2N' = 1 - 4tCompute D':D = (1 + 2t)^2D' = 2(1 + 2t)(2) = 4(1 + 2t)So,f'(t) = [ (1 - 4t)(1 + 2t)^2 - (t - 2t^2)(4)(1 + 2t) ] / (1 + 2t)^4Factor out (1 + 2t):f'(t) = [ (1 + 2t) [ (1 - 4t)(1 + 2t) - 4(t - 2t^2) ] ] / (1 + 2t)^4Simplify:f'(t) = [ (1 - 4t)(1 + 2t) - 4(t - 2t^2) ] / (1 + 2t)^3Now, expand the numerator:First term: (1 - 4t)(1 + 2t) = 1(1) + 1(2t) - 4t(1) - 4t(2t) = 1 + 2t - 4t - 8t^2 = 1 - 2t - 8t^2Second term: -4(t - 2t^2) = -4t + 8t^2Combine both terms:1 - 2t - 8t^2 - 4t + 8t^2 = 1 - 6tSo, f'(t) = (1 - 6t) / (1 + 2t)^3Set f'(t) = 0:1 - 6t = 06t = 1t = 1/6So, critical point at t = 1/6.We need to check if this is a maximum.Compute second derivative or check the sign of f'(t) around t = 1/6.For t < 1/6, say t = 0.1:f'(0.1) = (1 - 0.6)/ (1 + 0.2)^3 = 0.4 / 1.728 ≈ positiveFor t > 1/6, say t = 0.2:f'(0.2) = (1 - 1.2)/ (1 + 0.4)^3 = (-0.2)/ (1.4)^3 ≈ negativeSo, f(t) increases before t = 1/6 and decreases after, so t = 1/6 is a maximum.Therefore, the maximum area occurs at t = 1/6.Now, compute A(t) at t = 1/6:First, compute f(t):f(1/6) = (1/6)(1 - 2*(1/6)) / (1 + 2*(1/6))^2Simplify:1 - 2*(1/6) = 1 - 1/3 = 2/31 + 2*(1/6) = 1 + 1/3 = 4/3So,f(1/6) = (1/6)(2/3) / (4/3)^2 = (2/18) / (16/9) = (1/9) / (16/9) = 1/16Therefore, A^2(t) = 2 * f(t) = 2 * (1/16) = 1/8So, A(t) = sqrt(1/8) = (1)/(2√2) = √2 / 4Therefore, the maximum area is √2 / 4.So, the maximum area of triangle MNF is √2 / 4.Let me just verify the calculations.At t = 1/6:m^2 = 1/6, so m = 1/√6Then, the area:A = (m √2 √(1 - 2m^2)) / (1 + 2m^2)Compute 1 - 2m^2 = 1 - 2*(1/6) = 1 - 1/3 = 2/3So,A = ( (1/√6) * √2 * √(2/3) ) / (1 + 2*(1/6)) Simplify numerator:(1/√6) * √2 * √(2/3) = (1/√6) * √(4/3) = (1/√6) * (2/√3) = (2)/(√6 * √3) = 2 / √18 = 2 / (3√2) = √2 / 3Denominator:1 + 2*(1/6) = 1 + 1/3 = 4/3So,A = (√2 / 3) / (4/3) = √2 / 4Yes, that matches. So, the maximum area is indeed √2 / 4.Therefore, the answers are:(I) The standard equation is (frac{x^2}{2} + y^2 = 1).(II) (i) Proved that (angle AFM = angle BFN).(ii) The maximum area of triangle MNF is (frac{sqrt{2}}{4}).

💡Okay, so I have this problem about vectors, and I need to find the magnitude of the vector difference between two vectors, a and b. The given information is that the magnitude of vector a is 2, the magnitude of vector b is 3, and the angle between them is 60 degrees. The options are A: sqrt(19), B: 19, C: sqrt(7), and D: 7.Alright, let me start by recalling what I know about vectors and their magnitudes. I remember that when you have two vectors, the magnitude of their difference can be found using the formula that involves the dot product. The formula is something like |a - b| squared equals |a| squared plus |b| squared minus 2 times the dot product of a and b. So, writing that out, it would be |a - b|² = |a|² + |b|² - 2(a · b). I think that's right. Let me double-check. Yeah, that seems familiar from the law of cosines, which makes sense because when you subtract vectors, you're essentially forming a triangle, and the law of cosines relates the sides and the angle between them.Now, I need to compute the dot product of a and b. The dot product formula is a · b = |a||b|cos(theta), where theta is the angle between them. Given that |a| is 2, |b| is 3, and theta is 60 degrees, I can plug those values in. So, a · b = 2 * 3 * cos(60°). Wait, what's cos(60°)? I remember that cos(60°) is 0.5 or 1/2. So, a · b = 2 * 3 * 0.5. Let me calculate that: 2 times 3 is 6, and 6 times 0.5 is 3. So, the dot product is 3.Now, going back to the formula for |a - b| squared. That would be |a| squared plus |b| squared minus 2 times the dot product. So, |a| squared is 2 squared, which is 4. |b| squared is 3 squared, which is 9. The dot product is 3, so 2 times that is 6.Putting it all together: |a - b|² = 4 + 9 - 6. Let me add those up: 4 plus 9 is 13, and 13 minus 6 is 7. So, |a - b| squared is 7. Therefore, to find |a - b|, I take the square root of 7. Hmm, sqrt(7) is one of the options, option C. So, I think that's the answer. But wait, let me make sure I didn't make any mistakes in my calculations. First, checking the dot product: 2 * 3 is 6, times cos(60°) which is 0.5, so 6 * 0.5 is indeed 3. That seems correct.Then, plugging into the formula: 4 + 9 is 13, minus 6 is 7. Square root of 7 is approximately 2.645, which is a reasonable magnitude given the vectors' lengths and the angle between them. I don't see any errors in my steps. So, I think I'm confident that the magnitude of the vector difference |a - b| is sqrt(7), which is option C.

💡Okay, so I have this sequence {a_n} where a_1 is 1, and for any positive integer n, a_{2n} equals a_n plus n. I need to find the value of a_{2^{100}}. Hmm, that seems a bit tricky, but let me try to break it down step by step.First, let's understand the given conditions. The sequence starts with a_1 = 1. Then, for any n, a_{2n} = a_n + n. So, if I know a_n, I can find a_{2n} by just adding n to it. That seems straightforward, but since I need to find a_{2^{100}}, which is a term far out in the sequence, I probably need to find a pattern or a formula that can help me compute it without calculating all the previous terms.Let me start by computing the first few terms to see if I can spot a pattern.We know that a_1 = 1.Now, let's compute a_2. Since 2 is 2*1, so n=1 here. Using the formula, a_2 = a_1 + 1 = 1 + 1 = 2.Next, let's compute a_4. 4 is 2*2, so n=2. Therefore, a_4 = a_2 + 2 = 2 + 2 = 4.Similarly, a_8 would be a_4 + 4 = 4 + 4 = 8.Wait a minute, that's interesting. a_1 = 1, a_2 = 2, a_4 = 4, a_8 = 8. It seems like a pattern where a_{2^k} = 2^k. Let me test this further.Compute a_16. 16 is 2*8, so n=8. Therefore, a_16 = a_8 + 8 = 8 + 8 = 16.Yes, a_16 is 16, which is 2^4. So, it seems like for each power of 2, the term a_{2^k} is equal to 2^k. So, if this pattern continues, then a_{2^{100}} should be 2^{100}.But wait, is this always true? Let me check another term to be sure.Compute a_32. 32 is 2*16, so n=16. Therefore, a_32 = a_16 + 16 = 16 + 16 = 32. Yep, that's 2^5.So, it seems consistent. Each time we double n, the term a_n doubles as well. So, for n being a power of 2, a_n is equal to n itself.But let me think about why this is happening. The recursive formula is a_{2n} = a_n + n. So, if a_n is n, then a_{2n} would be n + n = 2n, which is consistent with a_{2n} = 2n. So, if a_n = n for n being a power of 2, then a_{2n} will also be 2n, which is the next power of 2.Therefore, by induction, if a_{2^k} = 2^k, then a_{2^{k+1}} = a_{2^k} + 2^k = 2^k + 2^k = 2^{k+1}. So, this holds true for all k.Therefore, the general formula for a_{2^k} is 2^k. So, for k = 100, a_{2^{100}} = 2^{100}.Wait, but just to be thorough, let me try to write out the recursive steps for a few more terms to ensure that there isn't any hidden complexity.Starting from a_1 = 1.a_2 = a_1 + 1 = 1 + 1 = 2.a_4 = a_2 + 2 = 2 + 2 = 4.a_8 = a_4 + 4 = 4 + 4 = 8.a_16 = a_8 + 8 = 8 + 8 = 16.a_32 = a_16 + 16 = 16 + 16 = 32.a_64 = a_32 + 32 = 32 + 32 = 64.a_128 = a_64 + 64 = 64 + 64 = 128.This pattern is consistent. Each time, when n is a power of 2, a_n is equal to n. So, it seems that for any k, a_{2^k} = 2^k.Therefore, applying this to k = 100, we get a_{2^{100}} = 2^{100}.I think this makes sense because each time we apply the recursive formula, we're effectively doubling the previous term, which is exactly what happens when you have powers of 2. So, the sequence is growing exponentially in this case, matching the powers of 2.Just to make sure, let me think about how the sequence behaves for numbers that are not powers of 2. For example, a_3. Since 3 is not a power of 2, how would we compute a_3?Wait, the recursive formula only gives us a_{2n} in terms of a_n. It doesn't directly give us a formula for a_{2n+1} or any odd terms. So, actually, we don't have information about the odd terms beyond a_1. But in our problem, we're only asked about a_{2^{100}}, which is a power of 2, so we don't need to worry about the odd terms here.Therefore, focusing solely on the powers of 2, the recursive formula gives us a straightforward way to compute a_{2^{100}} as 2^{100}.So, after all this thinking and checking, I'm pretty confident that the value of a_{2^{100}} is 2^{100}.

💡Alright, so I have this problem here about sets and a new operation defined for them. Let me try to wrap my head around it step by step. Okay, the operation is denoted by ×, but it's not the usual Cartesian product. Instead, it's defined as M × N = {x | x ∈ M or x ∈ N, but x ∉ M ∩ N}. Hmm, so this operation includes elements that are in either M or N, but not in both. That sounds familiar—it's like the symmetric difference of two sets. Yeah, I think that's right. The symmetric difference includes elements that are in either of the sets but not in their intersection.Alright, so given that, let's look at the sets M and N provided. M is {0, 2, 4, 6, 8, 10} and N is {0, 3, 6, 9, 12, 15}. The question asks for (M × N) × M. So, first, I need to compute M × N, and then take that result and perform the × operation with M again.Let me start by finding M × N. As per the definition, this should be all elements that are in M or N but not in both. So, first, I need to find the intersection of M and N, which is M ∩ N. Looking at M and N:M = {0, 2, 4, 6, 8, 10}N = {0, 3, 6, 9, 12, 15}The common elements are 0 and 6. So, M ∩ N = {0, 6}.Now, M × N should include all elements from M and N except those in the intersection. So, from M, we have elements {2, 4, 8, 10} (excluding 0 and 6), and from N, we have {3, 9, 12, 15} (excluding 0 and 6). So, combining these, M × N should be {2, 4, 8, 10, 3, 9, 12, 15}.Let me double-check that. M has 0, 2, 4, 6, 8, 10. N has 0, 3, 6, 9, 12, 15. The intersection is 0 and 6, so we exclude those. So, from M, we take 2, 4, 8, 10, and from N, we take 3, 9, 12, 15. Yep, that seems right.Now, the next step is to compute (M × N) × M. So, we have M × N = {2, 4, 8, 10, 3, 9, 12, 15}, and we need to perform the × operation with M again.First, let's find the intersection of (M × N) and M. So, (M × N) is {2, 4, 8, 10, 3, 9, 12, 15}, and M is {0, 2, 4, 6, 8, 10}. The common elements between these two sets are 2, 4, 8, 10. So, (M × N) ∩ M = {2, 4, 8, 10}.Now, according to the operation ×, (M × N) × M should include all elements that are in (M × N) or in M, but not in both. So, let's break it down.From (M × N), we have elements {2, 4, 8, 10, 3, 9, 12, 15}. From M, we have {0, 2, 4, 6, 8, 10}. The intersection is {2, 4, 8, 10}, so we need to exclude these elements from the union of (M × N) and M.So, the union of (M × N) and M is {0, 2, 3, 4, 6, 8, 9, 10, 12, 15}. Now, excluding the intersection {2, 4, 8, 10}, we get {0, 3, 6, 9, 12, 15}.Wait a minute, that's interesting. So, (M × N) × M results in {0, 3, 6, 9, 12, 15}, which is exactly the set N. Is that a coincidence? Let me think.Actually, no, it's not a coincidence. The operation × is the symmetric difference, and one of the properties of symmetric difference is that (A × B) × A = B. Let me verify that.If we take (A × B) × A, it's equivalent to (A Δ B) Δ A, which simplifies to B, because the symmetric difference is associative and commutative, and A Δ A is the empty set. So, (A Δ B) Δ A = B Δ (A Δ A) = B Δ ∅ = B.So, in this case, (M × N) × M = N. That's a neat property.But just to make sure, let's go through it again step by step.First, M × N = {2, 4, 8, 10, 3, 9, 12, 15}.Then, (M × N) × M: we need to find elements that are in either (M × N) or M, but not in both.So, (M × N) has {2, 4, 8, 10, 3, 9, 12, 15}, and M has {0, 2, 4, 6, 8, 10}.The union is {0, 2, 3, 4, 6, 8, 9, 10, 12, 15}.The intersection is {2, 4, 8, 10}.So, excluding the intersection from the union, we get {0, 3, 6, 9, 12, 15}, which is exactly N.Therefore, (M × N) × M = N.That makes sense because of the properties of symmetric difference. So, in general, (A × B) × A = B.But just to be thorough, let's consider another example to see if this holds.Suppose A = {1, 2, 3} and B = {3, 4, 5}.Then, A × B = {1, 2, 4, 5}.Now, (A × B) × A = {1, 2, 4, 5} × {1, 2, 3}.The intersection is {1, 2}.So, the symmetric difference would be {4, 5, 3}.But B is {3, 4, 5}, so it's the same as B.Hence, it seems that indeed (A × B) × A = B.Therefore, in our original problem, (M × N) × M = N.So, the final answer should be N, which is {0, 3, 6, 9, 12, 15}.But wait, in my initial calculation, I thought (M × N) × M was {0, 3, 6, 9, 12, 15}, which is N. So, that aligns with the property.But just to make sure, let's think about it differently.Suppose we define × as symmetric difference, which is A Δ B = (A ∪ B) (A ∩ B).Then, (A Δ B) Δ A = A Δ (A Δ B) = B, because Δ is commutative and associative.So, yes, (A Δ B) Δ A = B.Therefore, in our case, (M × N) × M = N.So, the answer is N, which is {0, 3, 6, 9, 12, 15}.But let me double-check my initial step-by-step to ensure I didn't make any mistakes.First, M × N:M = {0, 2, 4, 6, 8, 10}N = {0, 3, 6, 9, 12, 15}Intersection M ∩ N = {0, 6}So, M × N = (M ∪ N) (M ∩ N) = {0, 2, 3, 4, 6, 8, 9, 10, 12, 15} {0, 6} = {2, 3, 4, 8, 9, 10, 12, 15}Wait, in my initial calculation, I had {2, 4, 8, 10, 3, 9, 12, 15}, which is the same as {2, 3, 4, 8, 9, 10, 12, 15}. So, that's correct.Then, (M × N) × M:(M × N) = {2, 3, 4, 8, 9, 10, 12, 15}M = {0, 2, 4, 6, 8, 10}Intersection = {2, 4, 8, 10}So, (M × N) × M = ((M × N) ∪ M) ((M × N) ∩ M) = {0, 2, 3, 4, 6, 8, 9, 10, 12, 15} {2, 4, 8, 10} = {0, 3, 6, 9, 12, 15}, which is N.So, yes, that's correct.Therefore, the final answer is N, which is {0, 3, 6, 9, 12, 15}.But wait, in the problem statement, it says "then (M × N) × M equals ______." So, the answer is N, which is {0, 3, 6, 9, 12, 15}.But just to make sure, let's think about it in terms of elements.Starting with M × N, which is {2, 3, 4, 8, 9, 10, 12, 15}.Then, taking that set and performing × with M.So, elements in either set but not in both.From M × N: 2, 3, 4, 8, 9, 10, 12, 15.From M: 0, 2, 4, 6, 8, 10.So, elements in both are 2, 4, 8, 10.Therefore, the result is elements in M × N or M, but not both.So, from M × N: 3, 9, 12, 15.From M: 0, 6.So, combining these, we get {0, 3, 6, 9, 12, 15}, which is N.Yes, that's consistent.Therefore, the answer is N, which is {0, 3, 6, 9, 12, 15}.I think that's solid. I've double-checked it multiple ways, considering both the properties of symmetric difference and the step-by-step element comparison. So, I'm confident that (M × N) × M equals N.

💡First, I need to determine the number of handshakes between the two opposing teams. Each of the six members of one team shakes hands with each of the six members of the opposing team. This results in 6 multiplied by 6, which equals 36 handshakes.Next, I'll calculate the handshakes among the members of the same team. For one team, each of the six members shakes hands with the other five teammates. This gives 6 multiplied by 5, which equals 30 handshakes. However, since each handshake is counted twice in this calculation, I'll divide by 2 to get the actual number of unique handshakes, resulting in 15 handshakes per team. Since there are two teams, I'll multiply by 2, giving a total of 30 handshakes.Finally, I'll calculate the handshakes between all players and the referees. There are 12 players in total (6 from each team) and 3 referees. Each player shakes hands with each referee, so this results in 12 multiplied by 3, which equals 36 handshakes.Adding all these handshakes together: 36 (between teams) + 30 (within teams) + 36 (with referees) equals a total of 102 handshakes.

💡First, I need to determine the total number of towels Maria bought by adding the green and white towels together.Maria bought 40 green towels and 44 white towels, so the total number of towels is 40 + 44 = 84 towels.Next, Maria gave her mother 65 towels. To find out how many towels Maria has left, I subtract the number of towels she gave away from the total number of towels she originally had.84 towels - 65 towels = 19 towels.Therefore, Maria ended up with 19 towels.