Question

An ellipse intersects the hyperbola $2x^2-2y^2=1$ 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 the:

• A. equation of the ellipse is $x^2+2y^2=2$
• B. foci of the ellipse are$(\pm 1,0)$
• C. equation of the ellipse is $x^2+2y^2=4$
• D. foci of the ellipse are $(\pm \sqrt{2},0)$

Answer 1

Answer: (A), (B)

The correct options are
A foci of the ellipse are$(\pm 1,0)$
B equation of the ellipse is $x^2+2y^2=2$

Eccentricity of the hyperbola is $\sqrt{2}$ as it is a rectangular hyperbola so eccentricity $e$ of the ellipse is $\frac{1}{\sqrt{2}}$
Let the equation of the ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ where $b^2=a^2\left({1-e}^2\right)=\frac{a^2}{2}\Rightarrow a^2={2b}^2$
So equation of the ellipse is $x^2+{2y}^2=a^2$
Let $(x_1,y_1)$ be a point of the ellipse and the hyperbola
then ${2x}_1^2-{2y}_1^2=1$ and $x_1^2+{2y}_1^2=a^2$ ...(1)
Equations of the tangents at $(x_1,y_1)$ to the conics are ${2xx}_1-{2yy}_1=1$ and ${xx}_1+{2yy}_1=a^2$
Since the two conics intersect orthogonally $\left(\frac{x_1}{y_1}\right)(-\frac{x_1}{{2y}_1})=-1\Rightarrow x_1^2={2y}_1^2$
and from (1) we get $x_1^2=1,a^2=2$
Hence the equation of the ellipse is $x^2+{2y}^2=2$ and its focus is
$(\pm ae,0)=(\pm \sqrt{2}\times \frac{1}{\sqrt{2}},0)=(\pm 1,0)$