Most Important Selected Qs for JEE AdvancedMathematicsEllipse
An ellipse is orthogonal to the hyperbola x^2-y^2=2 . The eccentricity of the ellipse is reciprocal of that of the hyperbola. Then:
Options
- Aequation of the ellipse is x^2+2 y^2=8
- Bfocus of the ellipse is at (-4 2 , 0)
- Cequation of directrix of ellipse is x+4 2 =0
- Dequation of director circle of ellipse is x^2+y^2=12
Correct answer
D. equation of director circle of ellipse is x^2+y^2=12
Step-by-step solution
Eccentricity of ellipse = 1 2 Let equation of ellipse be x^2 a^2 + y^2 a^2 (1- 1 2 ) =1 x^2-y^2=2 As eqns. (1) and (2) intersect orthogonally, so . . d y d x ]_ (1) d y d x ]_ (2) =-1 at point of intersection.