NIMCET 2022 — Mathematics PYQ

First, let's rewrite the hyperbola equation in standard form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$:

Multiply the entire equation by 25:

Here, $a_h^2=\frac{144}{25}$ and $b_h^2=\frac{81}{25}$.

The eccentricity $e_h$ of a hyperbola is given by $e_h=\sqrt{1+\frac{b_h^2}{a_h^2}}$:

The foci of a hyperbola are $(\pm a_h{e}_h,0)$:

2. Analyze the Ellipse

The given ellipse is $\frac{x^2}{25}+\frac{y^2}{b^2}=1$.

Here, $a_e^2=25$, so $a_e=5$.

The foci of an ellipse are $(\pm a_e{e}_e,0)$. Since the foci coincide with the hyperbola, we have:

3. Calculate $b^2$

For an ellipse, the relationship between $a$, $b$, and $e$ is $b^2=a^2(1-e^2)$:

Final Answer:

The value of $b^2$ is 16. The correct option is B.