Hyperbola: A hyperbola passes through the foci of the ellipse… | JEE Main 2021

JEE Main 2021 (Online) 25th February Evening ShiftQuestion 1

A hyperbola passes through the foci of the ellipse $\frac{x^{2}}{25}+\frac{y^{2}}{16}=1$ and its transverse and conjugate axes coincide with major and minor axes of the ellipse, respectively. If the product of their eccentricities is one, then the equation of the hyperbola is :

Hyperbola

$\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$✓ Correct answer$\frac{x^{2}}{9}-\frac{y^{2}}{4}=1$$\frac{x^{2}}{9}-\frac{y^{2}}{25}=1$$x^{2}-y^{2}=9$

Solution

Ellipse: foci $(\pm3,0)$, $e_E=\tfrac35$, so $e_H=\tfrac53$. The hyperbola passes through the foci, which lie on its transverse axis, so $a=3$ and $b^2=9\left(\tfrac{25}{9}-1\right)=16$. $\dfrac{x^2}{9}-\dfrac{y^2}{16}=1$.

Hyperbola - JEE Mathematics question (JEE Main 2021 (Online) 25th February Evening Shift)

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