NIMCET 2012 — Mathematics PYQ

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Question

NIMCET 2012 — Mathematics PYQ

NIMCET | Mathematics | 2012

The equation of the ellipse with major axis along the $X$ -axis and passing through the points $(4,3)$ and $(-1,4)$ is:

Choose the correct answer:

Explanation

1. General Equation:

The standard equation of an ellipse with the major axis along the $X$ -axis is:

\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1

2. Substituting Point $(4,3)$ :

\frac{4^2}{a^2} + \frac{3^2}{b^2} = 1 \implies \frac{16}{a^2} + \frac{9}{b^2} = 1 \quad \dots \text{(Equation 1)}

3. Substituting Point $(-1,4)$ :

\frac{(-1)^2}{a^2} + \frac{4^2}{b^2} = 1 \implies \frac{1}{a^2} + \frac{16}{b^2} = 1 \quad \dots \text{(Equation 2)}

4. Solving for $\frac{1}{a^2}$ and $\frac{1}{b^2}$ :

Let $u = \frac{1}{a^2}$ and $v = \frac{1}{b^2}$ .

$16u + 9v = 1$
$u + 16v = 1 \implies u = 1 - 16v$

Substitute $u$ in the first equation:

16(1 - 16v) + 9v = 1

16 - 256v + 9v = 1

-247v = -15 \implies v = \frac{15}{247}

Now find $u$ :

u = 1 - 16\left(\frac{15}{247}\right) = \frac{247 - 240}{247} = \frac{7}{247}

5. Final Equation:

Substitute $u$ and $v$ back into $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ :

x^2\left(\frac{7}{247}\right) + y^2\left(\frac{15}{247}\right) = 1

7x^2 + 15y^2 = 247

Correct Option:

(b) 7x² + 15y² = 247

Choose the correct answer:

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