NIMCET 2014 — Mathematics PYQ

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Question

NIMCET 2014 — Mathematics PYQ

NIMCET | Mathematics | 2014

If the foci of the ellipse $b^{2}x^{2} + 16y^{2} = 16b^{2}$ and the hyperbola $81x^{2} - 144y^{2} = \frac{81 \times 144}{25}$ coincide, then the value of $b$ , is:

Choose the correct answer:

Explanation

Calculation for Hyperbola:

81x^{2} - 144y^{2} = \frac{81 \times 144}{25}

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

\therefore a_{h}^{2} = \frac{144}{25} \text{ and } b_{h}^{2} = \frac{81}{25}

e_{h}^{2} = 1 + \left( \frac{b_{h}^{2}}{a_{h}^{2}} \right)

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

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

F_{h} = (a_{h}e_{h}, 0) = \left( \frac{12}{5} \times \frac{15}{12}, 0 \right)

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

Calculation for Ellipse:

b^{2}x^{2} + 16y^{2} = 16b^{2}

\Rightarrow \frac{x^{2}}{16} + \frac{y^{2}}{b^{2}} = 1

\therefore a_{e}^{2} = 16 \text{ and } b_{e}^{2} = b^{2}

F_{e} = (a_{e}e_{e}, 0) = (4e_{e}, 0)

\text{Given: } F_{e} = F_{h}

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

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

e_{e}^{2} = 1 - \left( \frac{b_{e}^{2}}{a_{e}^{2}} \right)

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

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

\Rightarrow b^{2} = 7 \Rightarrow b = \sqrt{7}