NIMCET 2015 — Mathematics PYQ

1. Analyze the Hyperbola

First, let's write the hyperbola equation in standard form $\frac{x^2}{A^2}-\frac{y^2}{B^2}=1$.

The given equation is:

Multiply both sides by $25$:

Here, $A^2=\frac{144}{25}$ and $B^2=\frac{81}{25}$.

For a hyperbola, the distance of the focus from the center is $c_h$, where $c_h^2=A^2+B^2$:

So, $c_h=\sqrt{9}=3$.

The foci of the hyperbola are $(\pm 3,0)$.

2. Analyze the Ellipse

The ellipse equation is $\frac{x^2}{16}+\frac{y^2}{b^2}=1$.

Here, $a^2=16$.

The foci of the ellipse are $(\pm c_e,0)$. Since the foci coincide with those of the hyperbola:

For an ellipse, the relationship between the semi-axes and the focal distance is $c_e^2=a^2-b^2$ (assuming the horizontal axis is the major axis, which is consistent with the hyperbola foci being on the x-axis).

3. Solve for $b^2$

Substitute the known values into the ellipse formula:

Correct Option: (c)