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 value of $b^2$ is

• A. $1$
• B. $5$
• C. $7$
• D. $9$

Answer 1

Answer: (C)

$7$

Given ellipse: $\frac{x^2}{16}+\frac{y^2}{b^2}=1$

Now $b^2=a^2(1-e^2)$

$\Rightarrow b^2=16(1-e^2),\Rightarrow \frac{b^2}{16}=1-e^2$

$\Rightarrow e^2=1-\frac{b^2}{16}=\frac{16-b^2}{16}\Rightarrow e=\frac{\sqrt{16-b^2}}{4}$

$Foci=(\pm ae,0)=(\pm \sqrt{16-b^2},0)$

Given hyperbola: $\frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$

$\Rightarrow \frac{x^2}{{\left(\frac{12}{5}\right)}^2}-\frac{y^2}{{\left(\frac{9}{5}\right)}^2}=1$

Now, $b^2=a^2(e^2-1)$

$\Rightarrow {\left(\frac{9}{5}\right)}^2={\left(\frac{12}{5}\right)}^2(e^2-1)$

$\Rightarrow {\left(\frac{9}{12}\right)}^2=e^2-1$,

$\Rightarrow e^2=1+\frac{81}{144}=\frac{144+81}{144}$

$\Rightarrow e=\frac{15}{12}=\frac{5}{4}$

$Foci=(\pm ae,0)=(\pm 3,0)$

Since foci of the given ellipse and hyperbola coincide, therefore

$\sqrt{16-b^2}=3\Rightarrow 16-b^2=9$

$\therefore b^2=7$

Hence, option C is correct.