Question

The foci of the ellipse $\frac{{\mathrm{x}}^2}{16}+\frac{{\mathrm{y}}^2}{b^2}=1$ and the hyperbola $\frac{{\mathrm{x}}^2}{144}-\frac{{\mathrm{y}}^2}{81}=\frac{1}{25}$ coincide. Then the value of $b^2$ is

• A. $1$
• B. $5$
• C. $7$
• D. $9$

Answer 1

The correct option is C

$7$

Explanation for correct option:

Step-1 : Formula for eccentricity for both curves:

Given: The foci of the ellipse $\frac{{\mathrm{x}}^2}{16}+\frac{{\mathrm{y}}^2}{b^2}=1$ and the hyperbola $\frac{{\mathrm{x}}^2}{144}-\frac{{\mathrm{y}}^2}{81}=\frac{1}{25}$ coincide.

We know that ,

Formula for the eccentricity of hyperbola : $e=\sqrt{1+\frac{b^2}{a^2}}$

Formula for the eccentricity of ellipse : $e=\sqrt{1-\frac{b^2}{a^2}}$

Step-2 : Find the foci of the given conic section :

Hyperbola equation becomes,

$\frac{{\mathrm{x}}^2}{144}-\frac{{\mathrm{y}}^2}{81}=\frac{1}{25}$

$\Rightarrow \frac{{\mathrm{x}}^2}{{\displaystyle \frac{144}{25}}}-\frac{{\mathrm{y}}^2}{{\displaystyle \frac{81}{25}}}=1......\left(1\right)$

Now ,Comparing the equation (1) with the standard equation of hyperbola $\frac{{\mathrm{x}}^2}{a^2}-\hspace{0.17em}\frac{{\mathrm{y}}^2}{b^2}=1$ .

So , we have $a^2=\frac{144}{25}\Rightarrow a=\frac{12}{5},b^2=\frac{81}{25}\Rightarrow b=\frac{9}{5}$

Now, Eccentricity of the hyperbola :-

Assume $e$ is the eccentricity of hyperbola.

$$\begin{gathered} e=\sqrt{1+\frac{b^2}{a^2}} \\ \Rightarrow e=\sqrt{1+\frac{81}{144}} \\ \Rightarrow e=\sqrt{\frac{144+81}{144}} \\ \Rightarrow e=\sqrt{\frac{225}{144}} \\ \Rightarrow e=\frac{15}{12} \\ \Rightarrow e=\frac{5}{4} \end{gathered}$$

We know that , coordinates of foci of standard hyperbola are $(\pm ae,0)$

Here , $a=\frac{12}{5},e=\frac{5}{4}$

So , foci is $(\pm \frac{12}{5}\times \frac{5}{4}).$

Foci is $(\pm 3,0)$.

Step-3 : Equating foci of the both curves:

Now , Assume $e^{\text{'}}$ is the eccentricity of ellipse.

Therefore the foci of the ellipse and hyperbola are at same points.

$(\pm 4e,0)=(\pm \frac{12}{5}e\text{'},0)$ $\left[\because \mathrm{for}\mathrm{ellipse}a=4\right]$

$\Rightarrow$ $4e=\frac{12}{5}e’$

$\Rightarrow$$\frac{4\sqrt{16-{\mathrm{b}}^2}}{4}=\frac{12}{5}\sqrt{\frac{{\displaystyle \frac{144}{25}+\frac{81}{25}}}{{\displaystyle \frac{144}{25}}}}$

$\Rightarrow$ $\sqrt{16-{\mathrm{b}}^2}=\frac{12}{5}\frac{\sqrt{{\displaystyle \frac{225}{25}}}}{{\displaystyle \frac{12}{5}}}$

So, $$\begin{gathered} 16–b^2=9 \\ {b}^2=7 \end{gathered}$$

Hence, the value of $b^2$ is $7$,

Therefore, option (C) is the correct answer.