NIMCET 2016 — Mathematics PYQ

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Question

NIMCET 2016 — Mathematics PYQ

NIMCET | Mathematics | 2016

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

Concept:
The eccentricity of the curve $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1$ is $e^{2} = 1 - (\frac{b ^{2}}{a ^{2}})$
The eccentricity of the curve $\frac{x ^{2}}{a ^{2}} - \frac{y ^{2}}{b ^{2}} = 1$ is $e^{2} = 1 + (\frac{b ^{2}}{a ^{2}})$
Foci of Hyperbola and Ellipse are (ae, 0) and (4e, 0)

Calculation:

For given Hyperbola $\frac{x ^{2}}{144} - \frac{y ^{2}}{81} = \frac{1}{25}$

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

$∴ a_{h}^{2} = \frac{144}{25}$ and $b_{h}^{2} = \frac{81}{25}$

Eccentricity of hyperbola

$e_{h}^{2} = 1 + (\frac{b _{h}^{2}}{a _{h}^{2}})$

$\Rightarrow e_{h}^{2} = 1 + (\frac{81}{144})$

$\Rightarrow e_{h}^{2} = \frac{225}{144} \Rightarrow e_{h} = \frac{15}{12}$

Focus of hyperbola $F_{h} = (a_{h} e_{h}, 0)$ , Where $e_{h}$ is the eccentricity of the hyperbola.

$\Rightarrow F_{h} = ((\frac{12}{5} \times \frac{15}{12}), 0)$

$\Rightarrow F_{h} = (3, 0)$

For given Ellipse $\frac{x ^{2}}{16} + \frac{y ^{2}}{b ^{2}} = 1$

$∴ a_{e}^{2} = 16$ and $b_{e}^{2} = b^{2}$

Focus of ellipse $F_{e} = (a e, 0) = (4 e_{e}, 0)$ , Where $e_{e}$ is eccentricity of the ellipse.

Given Focus of ellipse $F_{e} = F_{h}$

$\Rightarrow (4 e_{e}, 0) = (3, 0)$

$\Rightarrow 4 e_{e} = 3 \Rightarrow e_{e} = \frac{3}{4}$

Also Eccentricity of an ellipse

$e_{e}^{2} = 1 - (\frac{b _{e}^{2}}{a _{e}^{2}})$

$\Rightarrow (\frac{3}{4})^{2} = 1 - (\frac{b ^{2}}{4 ^{2}})$

$\Rightarrow 1 - \frac{9}{16} = \frac{b ^{2}}{16}$

$\Rightarrow b^{2} = 7$