Let the eccentricity of an ellipse [x^2a^2+y^2b^2=1] is reciprocal to that of the hyperbol…

Question

Let the eccentricity of an ellipse $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$$ is reciprocal to that of the hyperbola $2 x^{2}-2 y^{2}=1$. If the ellipse intersects the hyperbola at right angles, then square of length of the latus-rectu...

Answer: 2

Accepted Answer

Equation of hyperbola is $2{x}^{2}-2{y}^{2}=1$

$\Rightarrow \frac{{x}^{2}}{1/2}-\frac{{y}^{2}}{1/2}=1$

Here, $a=b$

$	herefore $ Eccentricity of rectangular hyperbola is $\sqrt{2}$

$	herefore $ Eccentricity of ellipse is $\frac{1}{\sqrt{2}}$

Since, ellipse intersects the hyperbola at right angles

$	herefore $ Ellipse and the hyperbola are confocal

$	herefore $ Foci of hyperbola $=(1,0)$

$\Rightarrow$ Foci of ellipse, $ae=1$

$\begin{gathered} \Rightarrow a(\frac{1}{\sqrt{2}})=1 \ \Rightarrow a=\sqrt{2} \ 	herefore {e}^{2}=1-\frac{{b}^{2}}{{a}^{2}} \ \Rightarrow \frac{1}{2}=1-\frac{{b}^{2}}{2} \ \Rightarrow \frac{{b}^{2}}{2}=\frac{1}{2} \ \Rightarrow {b}^{2}=1 \end{gathered}$

$	herefore $ Length of the latus rectum $=\frac{2{b}^{2}}{a}=\frac{2}{\sqrt{2}}=\sqrt{2}$

$	herefore $ Square of length of the latus rectum $=2$

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