JEE PYQ: Hyperbola Question 5
Question 5 - 2021 (25 Feb Shift 2)
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:
(1) $\frac{x^2}{9} - \frac{y^2}{4} = 1$
(2) $\frac{x^2}{9} - \frac{y^2}{16} = 1$
(3) $x^2 - y^2 = 9$
(4) $\frac{x^2}{9} - \frac{y^2}{25} = 1$
Show Answer
Answer: (2)
Solution
Ellipse: $a^2 = 25$, $b^2 = 16$, $e_1 = \sqrt{1-\frac{16}{25}} = \frac{3}{5}$. Foci at $(\pm 3, 0)$. Let hyperbola be $\frac{x^2}{A^2} - \frac{y^2}{B^2} = 1$. Passes through $(\pm 3, 0)$: $A^2 = 9$, so $A = 3$. $e_1 \cdot e_2 = 1 \Rightarrow e_2 = \frac{5}{3}$. $B^2 = A^2(e_2^2 - 1) = 9 \cdot \frac{16}{9} = 16$. Equation: $\frac{x^2}{9} - \frac{y^2}{16} = 1$.
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