JEE Advance - Mathematics (2009 - Paper 2 Offline - No. 13)
An ellipse intersects the hyperbola $$2{x^2} - 2{y^2} = 1$$ orthogonally. The eccentricity of the ellipse is reciprocal of that of the hyperbola. If the axes of the ellipse are along the coordinate axes then
equation of ellipse is $${x^2} + 2{y^2} = 2$$
the foci of ellipse are $$\left( { \pm 1,0} \right)$$
equation of ellipse is $${x^2} + 2{y^2} = 4$$
the foci of ellipse are $$\left( { \pm \sqrt 2 ,0} \right)$$
Explanation
The ellipse and hyperbola will be confocal. Therefore,
$$( \pm ae,0) \equiv ( \pm 1,0)$$
$$ \Rightarrow \left( { \pm a \times {1 \over {\sqrt 2 }},0} \right) \equiv ( \pm 1,0)$$
$$ \Rightarrow a = \sqrt 2 $$ and $$e = {1 \over {\sqrt 2 }}$$
$$ \Rightarrow {b^2} = {a^2}(1 - {e^2}) \Rightarrow {b^2} = 1$$
Hence, the equation of ellipse $${{{x^2}} \over 2} + {{{y^2}} \over 1} = 1$$.
Comments (0)
