Question

If the focii of the ellipse $\frac{x^2}{25}+\frac{y^2}{b^2}=1$ and the hyperbola $\frac{x^2}{144}-\frac{y^2}{125}=\frac{1}{13}$ coincide, then the value of $b^2$ is

• A. $5$
• B. $12$
• C. $13$
• D. $17$
• E. $21$

Answer 1

Answer: (B)

$12$

Given equation of ellipse, $\frac{x^2}{25}+\frac{y^2}{b^2}=1$ .... (i)
and equation of hyperbola, $\frac{x^2}{144}-\frac{y^2}{24}=\frac{1}{13}$
$\Rightarrow \frac{x^2}{\left(\frac{144}{13}\right)}-\frac{y^2}{\left(\frac{25}{13}\right)}=1$
Here, $A^2=\frac{144}{33}$ and $B^2=\frac{25}{13}$
Eccentricity of hyperbola is,
$e_2=\sqrt{\frac{A^2+B^2}{A^2}}=\sqrt{\frac{169}{13}\times \frac{13}{144}}=\frac{13}{12}$
Focii of hyperbola is $(\pm A{e}_2,0)$
$=(\pm \frac{12}{\sqrt{13}}\times \frac{13}{12},0)=(\pm \sqrt{13},0)$
Given, focii of the ellipse and hyperbola coincide.
i.e., focii of an ellipse $(\pm a{e}_1,0)=(\pm \sqrt{13},0)$
$\Rightarrow a{e}_1=\sqrt{13}$
$\Rightarrow 5e_1=\sqrt{13}$
$\Rightarrow e_1=\frac{\sqrt{13}}{5}$
Also, $e_1=\sqrt{\frac{a^2-b^2}{a^2}}$
$\Rightarrow e_1^2{a}^2=a^2-b^2$
$\Rightarrow b^2=a^2-e_1^2{a}^2=25-\frac{13}{25}\times 25$
$\Rightarrow b^2=12$