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 $(\pm ae,0)=\left(\pm 13\times \frac{12}{13},0\right)$
$=(\pm 12,0)$
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
$e^{{}^{\prime }}=\sqrt{1+\frac{b^2}{a^2}}$
$=\sqrt{1+\frac{b^2}{144}}$
According to the equation,
$e{e}^{{}^{\prime }}=1$
$\frac{12}{13}\times \sqrt{1+\frac{b^2}{144}}=1$
$\Rightarrow \sqrt{1+\frac{b^2}{144}}=\frac{13}{12}$
$\Rightarrow 1+\frac{b^2}{144}=\frac{169}{144}$
$\Rightarrow \frac{b^2}{144}=\frac{169}{144}-1$
$\Rightarrow \frac{b^2}{144}=\frac{25}{144}$
$\Rightarrow b^2=25$
$\therefore$ Equation of hyperbola is
$\frac{x^2}{144}-\frac{y^2}{25}=1$