JEE MAIN - Mathematics (2023 - 6th April Evening Shift - No. 15)
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-rectum of the ellipse is ___________.
Answer
2
Explanation
Equation of hyperbola is $2 x^2-2 y^2=1$
$\Rightarrow \frac{x^2}{1 / 2}-\frac{y^2}{1 / 2}=1$
Here, $a=b$
$\therefore$ Eccentricity of rectangular hyperbola is $\sqrt{2}$
$\therefore$ Eccentricity of ellipse is $\frac{1}{\sqrt{2}}$
Since, ellipse intersects the hyperbola at right angles
$\therefore$ Ellipse and the hyperbola are confocal
$\therefore$ Foci of hyperbola $=(1,0)$
$\Rightarrow$ Foci of ellipse, $a e=1$
$$ \begin{array}{ll} &\Rightarrow a\left(\frac{1}{\sqrt{2}}\right)=1 \\\\ &\Rightarrow a=\sqrt{2} \\\\ &\therefore e^2=1-\frac{b^2}{a^2} \\\\ &\Rightarrow \frac{1}{2}=1-\frac{b^2}{2} \\\\ &\Rightarrow \frac{b^2}{2}=\frac{1}{2} \\\\ &\Rightarrow b^2=1 \end{array} $$
$\therefore$ Length of the latus rectum $=\frac{2 b^2}{a}=\frac{2}{\sqrt{2}}=\sqrt{2}$
$\therefore$ Square of length of the latus rectum $=2$
$\Rightarrow \frac{x^2}{1 / 2}-\frac{y^2}{1 / 2}=1$
Here, $a=b$
$\therefore$ Eccentricity of rectangular hyperbola is $\sqrt{2}$
$\therefore$ Eccentricity of ellipse is $\frac{1}{\sqrt{2}}$
Since, ellipse intersects the hyperbola at right angles
$\therefore$ Ellipse and the hyperbola are confocal
$\therefore$ Foci of hyperbola $=(1,0)$
$\Rightarrow$ Foci of ellipse, $a e=1$
$$ \begin{array}{ll} &\Rightarrow a\left(\frac{1}{\sqrt{2}}\right)=1 \\\\ &\Rightarrow a=\sqrt{2} \\\\ &\therefore e^2=1-\frac{b^2}{a^2} \\\\ &\Rightarrow \frac{1}{2}=1-\frac{b^2}{2} \\\\ &\Rightarrow \frac{b^2}{2}=\frac{1}{2} \\\\ &\Rightarrow b^2=1 \end{array} $$
$\therefore$ Length of the latus rectum $=\frac{2 b^2}{a}=\frac{2}{\sqrt{2}}=\sqrt{2}$
$\therefore$ Square of length of the latus rectum $=2$
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