NIMCET 2015 — Mathematics PYQ

Q: How do I solve NIMCET 2015 Mathematics questions?

Practice NIMCET 2015 Mathematics previous year questions with step-by-step solutions on Mathem Solvex. Each question includes a detailed explanation and video solution to help you master the concept quickly.

Q: What is the difficulty level of NIMCET Mathematics questions?

NIMCET Mathematics questions range from moderate to advanced difficulty. Consistent practice with official previous year papers and topic-wise drills on Mathem Solvex is the most effective preparation strategy.

Q: Are NIMCET previous year papers available for free?

Yes. Mathem Solvex provides free access to NIMCET previous year question papers and topic-wise PDFs. You can download them or attempt them as timed live mock tests directly on the platform.

Question

NIMCET 2015 — Mathematics PYQ

NIMCET | Mathematics | 2015

The radius of the circle passing through the foci of the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ and having its centre at $(0, 3)$ is:

Choose the correct answer:

Explanation

1. Find the properties of the Ellipse:

The given equation is $\frac{x^2}{16} + \frac{y^2}{9} = 1$ .

Comparing this with the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ :

$a^2 = 16 \implies a = 4$
$b^2 = 9 \implies b = 3$

Since $a > b$ , the major axis is along the X-axis.

2. Calculate the Eccentricity ( $e$ ):

The relationship between $a$ , $b$ , and $e$ is:

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

9 = 16(1 - e^2)

\frac{9}{16} = 1 - e^2

e^2 = 1 - \frac{9}{16} = \frac{7}{16}

e = \frac{\sqrt{7}}{4}

3. Find the Foci of the Ellipse:

The coordinates of the foci are $(\pm ae, 0)$ :

ae = 4 \times \frac{\sqrt{7}}{4} = \sqrt{7}

So, the foci are $S(\sqrt{7}, 0)$ and $S'(-\sqrt{7}, 0)$ .

4. Find the Radius of the Circle:

The circle has its centre at $C(0, 3)$ and passes through the foci $(\pm \sqrt{7}, 0)$ .

The radius ( $R$ ) is the distance between the centre $(0, 3)$ and one of the foci $(\sqrt{7}, 0)$ .

Using the distance formula:

R = \sqrt{(\sqrt{7} - 0)^2 + (0 - 3)^2}

R = \sqrt{(\sqrt{7})^2 + (-3)^2}

R = \sqrt{7 + 9}

R = \sqrt{16}

R = 4

Correct Option:

(a) $4$ units