Question

A hyperbola passing through a focus of the ellipse $\frac{x^2}{169}+\frac{y^2}{25}=1$.

Its transverse axis 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. $\frac{x^2}{25}-\frac{y^2}{9}=1$(x² / 25) – (y² / 9) = 1

Answer 1

The correct option is C

$\frac{x^2}{144}-\frac{y^2}{25}=1$

Explanation for the correct option:

Step 1. Find the equation of hyperbola:

For the given ellipse,

$\begin{array}{rcl}25& =& 169(1–e^2)\\ & \Rightarrow & e^2=1–\frac{25}{169}\\ & =& \frac{144}{169}\\ e& =& \left(\frac{12}{13}\right)\\ \therefore e& =& \frac{12}{13}\end{array}$

∴ The focus of the given ellipse

$\begin{array}{rcl}ae& =& 13e\\ & =& 13\frac{12}{13}\\ & =& 12\end{array}$

Step 2. Let the hyperbola be $\frac{x^2}{A^2}–\frac{y^2}{B^2}=1$

The hyperbola passes through $(12,0)$

$\begin{array}{rcl}\therefore \frac{{12}^2}{A^2}& =& 1\\ {12}^2& =& A^2\\ & \Rightarrow & A^2=144\end{array}$

Step 3. Let e₁ be the eccentricity of the hyperbola

$\begin{array}{rcl}{e}_1& =& \frac{1}{e}=\frac{13}{12}\\ \because B^2& =& A^2({e_1}^2-1)\\ & =& 144(\frac{169}{144}-1)\\ & =& 25\end{array}$

$\therefore$The required equation of hyperbola is $\frac{{\mathbf{x}}^{\mathbf{2}}}{\mathbf{144}}\mathbf{-}\frac{{\mathbf{y}}^{\mathbf{2}}}{\mathbf{25}}\mathbf{=}\mathbf{1}$

Hence, the correct option is $\left(C\right)$.