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 value of $b^2$ is:

• A. $5$
• B. $7$
• C. $9$
• D. $4$

Answer 1

Answer: (B)

$7$

The foci of the ellipse are also the foci of an hyperbola,
then we have, for the ellipse,

$a^2-c^2=b^2$

so

$16-c^2=b^2...............\left(1\right)$

Equation of Hyperbola can also be written as $\frac{x^2}{\frac{144}{25}}-\frac{y^2}{\frac{81}{25}}=1$

For the hyperbola, which must have its transverse axis on the x-axis, the equation
$c^2-a^2=b^2\Rightarrow c^2-\frac{144}{25}=\frac{81}{25}\Rightarrow c^2=\frac{225}{25}=9$

Putting this value in equation (1)

$16-9=b^2\Rightarrow b^2=7$