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UPSC NDA II 2015 Mathematics Paper

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Question : 32 of 120

Marks: +1, -0

The eccentricity of the hyperbola

16x^{2} - 9y^{2} = 1

is

$\frac{3}{5}$
$\frac{5}{3}$
$\frac{4}{5}$
$\frac{5}{4}$

Solution:

Concept:

For a hyperbola in standard form

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

, eccentricity

e

is given by

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

.

Explanation:

Step 1: Rewrite the given equation

16x^2 - 9y^2 = 1

in the form

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

.

Divide both sides by

1

:

\frac{x^2}{\frac{1}{16}} - \frac{y^2}{\frac{1}{9}} = 1

.

Step 2: Identify

a^2 = \frac{1}{16}

and

b^2 = \frac{1}{9}

.

Step 3: Substitute into the eccentricity formula:

e = \sqrt{1 + \frac{b^2}{a^2}} = \sqrt{1 + \frac{\frac{1}{9}}{\frac{1}{16}}} = \sqrt{1 + \frac{16}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3}

.

Answer:

\frac{5}{3}

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