Let an ellipse E₁ have its major axis along the x-axis, minor axis of length 6 , and latus rectum of length 9 2 . A second ellipse E₂ has identical dimensions to E₁ but its major axis lies along the y-axis. If E₁ and E₂ intersect at four points, and tangents are drawn to E₁ at these four points of intersection to form a quadrilateral Q , then the area of Q is :

Step-by-step solution

For E₁ , the length of the minor axis is 2b = 6 b = 3 b^2 = 9 . The length of the latus rectum is 2b^2 a = 9 2 . Substituting b^2 = 9 , we get 18 a = 9 2 a = 4 a^2 = 16 . The equations of the ellipses are: E₁: x^2 16 + y^2 9 = 1 E₂: x^2 9 + y^2 16 = 1 To find the points of intersection, we solve the two equations simultaneously. By symmetry, x^2 = y^2 . x^2 16 + x^2 9 = 1 x^2 ( 25 144 ) = 1 x^2 = 144 25 The intersection point in the first quadrant is P ( 12 5 , 12 5 ) . The equation of the tangent to E₁ at P is: x