An ellipse intersects the hyperbola 2 x^2-2 y^2=1 orthogonally. The eccentricity of the ellipse is reciprocal to that of the hyperbola. If the axes of the ellipse are along the coordinate axes, then

Step-by-step solution

Given, 2 x^2-2 y^2=1 x^2 ( 1 2 ) - y^2 ( 1 2 ) =1 Eccentricity of hyperbola = 2 So, eccentricity of ellipse =1 / 2 Let equation of ellipse be aligned & x^2 a^2 + y^2 b^2 =1(a>b) 1 2 = 1- b^2 a^2 & b^2 a^2 = 1 2 a^2=2 b^2 & x^2+2 y^2=2 b^2 & aligned Let ellipse and hyperbola intersect at A ( 1 2 , 1 2 ) On differentiating Eq. (i), 4 x-4 y d y d x =0 d y d x = x y ( d y d x )_ at A = = cosec On differentiating Eq. (ii), 2 x+4 y d y d x =0 ( d y d x )_ at A =- x 2 y =- 1 2 cosec Since, ellipse and hyperbola are orthog