A hyperbola passing through a focus of the ellipse $\frac{x^2}{169}+\frac{y^2}{25}=1$. Its transverse and…
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
$\frac{x^2}{144}-\frac{y^2}{9}=1$
$\frac{x^2}{169}-\frac{y^2}{25}=1$
$\frac{x^2}{144}-\frac{y^2}{25}=1$
$\frac{x^2}{25}-\frac{y^2}{9}=1$
Solution
Let the equation of hyperbola be
$
\frac{x^2}{a^2}-\frac{y^2}{b^2}=1
$
Given equation of ellipse is
$
\frac{x^2}{(13)^2}+\frac{y^2}{(5)^2}=1
$
Here,
$
\begin{aligned}
& \text { Here, } \quad a=13, b=5 \\
& \therefore \quad e=\sqrt{1-\frac{b^2}{a^2}} \\
& \therefore \quad=\sqrt{1-\frac{25}{169}}=\sqrt{\frac{144}{169}}=\frac{12}{13} \\
& \therefore \text { Focus }( \pm a e, 0)=\left( \pm 13 \times \frac{12}{13}, 0\right) \\
& \quad=( \pm 12,0)
\end{aligned}
$
Since, Eq. (i) passes through $( \pm 12,0)$.
$
\begin{array}{rrrl}
\therefore & \frac{144}{a^2}-\frac{0}{b^2}=1 \\
\Rightarrow & a^2=144 \\
\Rightarrow & a= \pm 12
\end{array}
$
Now eccentricity of hyperbola
$
\begin{aligned}
e^{\prime} & =\sqrt{1+\frac{b^2}{a^2}} \\
& =\sqrt{1+\frac{b^2}{144}}
\end{aligned}
$
According to the equation,
$
\begin{array}{rlrl}
& e^{\prime}=1 \\
\Rightarrow & & \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} \\
b^2 & =25
\end{array}
$
$\therefore$ Equation of hyperbola is
$
\frac{x^2}{144}-\frac{y^2}{25}=1
$