Question

A hyperbola passes 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

The correct option is C $\frac{x^2}{144}-\frac{y^2}{25}=1$

The equation of the ellipse can be written as $\frac{x^2}{{13}^2}+\frac{y^2}{5^2}=1$
Hence, $a=13$ and $b=5$
$e=\sqrt{1-\frac{b^2}{a^2}}$

$=\sqrt{\frac{a^2-b^2}{a^2}}$
$=\sqrt{\frac{169-25}{169}}$
$=\frac{12}{13}$
Now let the eccentricity of the hyperbola be $e^{\prime }$
$e.e^{\prime }=1$
$e^{\prime }=\frac{1}{e}=\frac{13}{12}$
$\sqrt{1+\frac{b^{\prime 2}}{a^{\prime 2}}}=\frac{13}{12}$
$\sqrt{\frac{a^{\prime 2}+b^{\prime 2}}{a^{\prime 2}}}=\frac{13}{12}$
$\frac{\sqrt{a^{\prime 2}+b^{\prime 2}}}{a^{\prime }}=\frac{13}{12}$

$a^{\prime }=12$ and $a^{\prime 2}+b^{\prime 2}=169$
$b^{\prime 2}=169-a^{\prime 2}$ $=169-144$ $=25$
Therefore $b^{\prime }=5$
Hence, the equation of the hyperbola is $\frac{x^2}{144}-\frac{y^2}{25}=1$