Question

The foci of the ellipse $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ and the hyperbola $\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$ coincide, then the sum of the squares of length of major axis and minor axis of the ellipse is

• A. 23
• B. 78
• C. 92
• D. 11

Answer 1

The correct option is C 92

Equation of hyperbola is $\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$
$\Rightarrow \frac{x^2}{\left(\frac{144}{25}\right)}-\frac{y^2}{\left(\frac{81}{25}\right)}=1$
$foci=(\pm \sqrt{a^2+b^2},0)=(\pm 3,0)$
foci of ellipse = $\frac{x^2}{16}+\frac{y^2}{b^2}=1$ (4>b)
$=(\pm \sqrt{16-b^2},0)=(\pm 3,0)$
$\Rightarrow b^2=7\Rightarrow b=\sqrt{7},a=4$
Required sum $=8^2+(2\sqrt{7}{)}^2=64+28=92$