Question

A hyperbola passing through a focus of the ellipse $\frac{x^{2}}{169}+\frac{y^{2}}{25}=1 .$ Its transverse and conjugate axes coincide respectively with the major and minor axes of the ellipse. The product of eccentricities is $1 $ . Then, the equation of the hyperbola is

• A. $\frac{x^{2}}{144}-\frac{y^{2}}{9}=1$
• B. $\frac{x^{2}}{169}-\frac{y^{2}}{25}=1$
• C. $\frac{x^{2}}{144}-\frac{y^{2}}{25}=1$
• D. $\frac{x^{2}}{25}-\frac{y^{2}}{9}=1$

Answer 1

Answer: (C)

Let the equation of hyperbola be
$\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\,\,\,...(i)$
Given equation of ellipse is
$\frac{x^{2}}{(13)^{2}}+\frac{y^{2}}{(5)^{2}}=1 $
Here $a=13, b=5$
$\therefore e=\sqrt{1-\frac{b^{2}}{a^{2}}}$
$=\sqrt{1-\frac{25}{169}}=\sqrt{\frac{144}{169}}=\frac{12}{13}$
$\therefore $ Focus $(\pm a e, 0) =\left(\pm 13 \times \frac{12}{13}, 0\right) $
$=(\pm 12,0)$
Since, Eq. (i) passes through $(\pm 12,0)$.
$\therefore \frac{144}{a^{2}}-\frac{0}{b^{2}}=1$
$\Rightarrow a^{2}=144$
$\Rightarrow a=\pm 12$
Now eccentricity of hyperbola
$e^{'}=\sqrt{1+\frac{b^{2}}{a^{2}}}$
$=\sqrt{1+\frac{b^{2}}{144}}$
According to the equation,
$e e^{'}=1$
$\frac{12}{13} \times \sqrt{1+\frac{b^{2}}{144}}=1$
$\Rightarrow \sqrt{1+\frac{b^{2}}{144}}=\frac{13}{12}$
$\Rightarrow 1+\frac{b^{2}}{144}=\frac{169}{144}$
$\Rightarrow \frac{b^{2}}{144}=\frac{169}{144}-1$
$\Rightarrow \frac{b^{2}}{144}=\frac{25}{144}$
$\Rightarrow b^{2}=25$
$\therefore $ Equation of hyperbola is
$\frac{x^{2}}{144}-\frac{y^{2}}{25}=1$