Passage: Consider the ellipses given by x^2 + 4y^2 = 1 and 4x^2 + y^2 = 1. Question: Let P be the point in the first quadrant where the given ellipses intersect. If is the acute angle between the tangents to the given ellipses at the point P, then the value of 4 is ___________.
A. A
The given equations of the ellipses are x^2 + 4y^2 = 1 and 4x^2 + y^2 = 1. To find the point of intersection, we equate the two expressions: x^2 + 4y^2 = 4x^2 + y^2 3x^2 = 3y^2 x = y (since P lies in the first quadrant) Substituting x = y into the first ellipse equation: x^2 + 4x^2 = 1 5x^2 = 1 x = 1 5 Thus, the point of intersection is P ( 1 5 , 1 5 ). Differentiating x^2 + 4y^2 = 1 with respect to x gives the slope of the tangent to the first ellipse: 2x + 8y dy dx = 0 dy dx = - x 4y At P ( 1 5 , 1 5 ), the slope is m_1 = - 1 4 . Differentiating 4x^2 + y^2 = 1 with respect to x gives the slope of the tangent to the second ellipse: 8x + 2y dy dx = 0 dy dx = - 4x y At P ( 1 5 , 1 5 ), the slope is m_2 = -4. The acute angle between the tangents is given by: = | m_1 - m_2 1 + m_1 m_2 | = | - 1 4 - (-4) 1 + (- 1 4 )(-4) | = | 15 4 2 | = 15 8 Therefore, the value of 4 is: 4 = 4 15 8 = 15 2 = 7.5 Answer: 7.5
Related: Mathematics — Ellipse · All PYQ Banks