NIMCET 2011 — Mathematics PYQ

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Question

NIMCET 2011 — Mathematics PYQ

NIMCET | Mathematics | 2011

The eccentricity of ellipse $9x^2 + 5y^2 - 30y = 0$ is:

Choose the correct answer:

Explanation

1. Complete the Square for the y-terms:

The given equation is:

9x^2 + 5(y^2 - 6y) = 0

To complete the square inside the parentheses, add $(\frac{-6}{2})^2 = 9$ :

9x^2 + 5(y^2 - 6y + 9) = 5(9)

9x^2 + 5(y - 3)^2 = 45

2. Convert to Standard Form:

Divide the entire equation by 45:

\frac{9x^2}{45} + \frac{5(y - 3)^2}{45} = \frac{45}{45}

\frac{x^2}{5} + \frac{(y - 3)^2}{9} = 1

3. Identify Semi-axes:

Comparing with $\frac{x^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$ :

$a^2 = 5$
$b^2 = 9$

Since $b^2 > a^2$ , this is a vertical ellipse.

4. Calculate Eccentricity ( $e$ ):

For an ellipse where $b > a$ , the eccentricity is given by the formula:

e = \sqrt{1 - \frac{a^2}{b^2}}

Substitute the values:

e = \sqrt{1 - \frac{5}{9}}

e = \sqrt{\frac{9 - 5}{9}}

e = \sqrt{\frac{4}{9}}

e = \frac{2}{3}

Correct Option: (b)