NIMCET 2015 — Mathematics PYQ

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Question

NIMCET 2015 — Mathematics PYQ

NIMCET | Mathematics | 2015

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:

Choose the correct answer:

Explanation

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:

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

Multiply both sides by $25$ :

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

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 $c_h^2 = A^2 + B^2$ :

c_h^2 = \frac{144}{25} + \frac{81}{25} = \frac{225}{25} = 9

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:

c_e = 3

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:

3^2 = 16 - b^2

9 = 16 - b^2

b^2 = 16 - 9

b^2 = 7

Correct Option: (c)