Question

If a hyperbola passes through the foci of the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ and its traverse and conjugate axis coincide with major and minor axes of the ellipse, and product of the eccentricities is 1, then:

• A. Equations of the hyperbola is $\frac{x^2}{9}-\frac{y^2}{16}=1$
• B. Equations of the hyperbola is $\frac{x^2}{9}-\frac{y^2}{25}=1$
• C. Focus of the hyperbola is $(5,0)$
• D. Focus of the hyperbola is $(5\sqrt{3},0)$

Answer 1

Answer: (A), (C)

The correct options are
A Equations of the hyperbola is $\frac{x^2}{9}-\frac{y^2}{16}=1$
C Focus of the hyperbola is $(5,0)$

Given ellipse is, $\frac{x^2}{25}+\frac{y^2}{16}=1$
Eccentricity of the ellipse is, $e_e=\sqrt{1-\frac{b^2}{a^2}}=\frac{3}{5}$
So the foci of the ellipse is, $(\pm a{e}_e,0)=(\pm 3,0)$
Let eccentricity of the required hyperbola is $e_h$ and semi major and minor axes are $a$ and $b$, so the equation of hyperbola is, $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
Given hyperbola passes trough $(\pm 3,0)\Rightarrow \frac{9}{a^2}-\frac{0}{b^2}=1\Rightarrow a^2=9$
Also given that $e_e\times e_h=1$ $\Rightarrow e_h=\frac{5}{3}$ $\Rightarrow$ $b^2=a^2(e_h^2-1)=16$
Hence required hyperbola is, $\frac{x^2}{9}-\frac{y^2}{16}=1$
And foci of the hyperbola is, $(\pm 5,0)$