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 coordinates axes, then

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

Answer 1

Answer: (A), (B)

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

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 $b^2=a^2(1-e^2)=\frac{a^2}{2}\Rightarrow a^2=2b^2$
So equation of the ellipse is $x^2+2y^2=a^2$
Let $(x_1,y_1)$ be a point of intersection 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 two conics are
$2x{x}_1-2y{y}_1=1$ and $x{x}_1+2y{y}_1=a^2$
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
$(\pm ae,0)=(\pm \sqrt{2}\times \frac{1}{\sqrt{2}},0)=(\pm 1,0)$