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KEAM 2014 Math Paper

Question : 120 of 120

Marks: +1, -0

2\cos \alpha.

9x^2+16y^2=144.

The equation of the hyperbola is

Solution:

Let equation of hyperbola is

\;\frac{x^{2}}{a_{1}^{2}}-\;\frac{y^{2}}{b_{1}^{2}}=1

Given,

\;\; 2 a_{1}=2 \cos \alpha

\Rightarrow \;\; a_{1}=\cos \alpha

Also, given equation of ellipse is

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

Here,

e=\sqrt{1-\;\frac{b^{2}}{a^{2}}}=\sqrt{1-\;\frac{9}{16}}=\;\frac{\sqrt{7}}{4}

According to the given condition,

Foci of hyperbola

(e_{1})=

Foci of ellipse

(e)

\Rightarrow \;\; \pm a_{1} e_{1}=\pm a e

\Rightarrow \;\; \cos \alpha \cdot e_{1}=4 \cdot \;\frac{\sqrt{7}}{4}

\Rightarrow \;\; \cos \alpha \cdot e_{1}=\sqrt{7}

\Rightarrow \cos \alpha \sqrt{1+\;\frac{b_{1}^{2}}{\cos^{2} \alpha}}=\sqrt{7}

\Rightarrow \;\; \cos^{2} \alpha+b_{1}^{2}=7

\Rightarrow \;\; b_{1}^{2}=7-\cos^{2} \alpha

\therefore

The equation of hyperbola is

\;\frac{x^{2}}{\cos^{2} \alpha}-\;\frac{y^{2}}{7-\cos^{2} \alpha}=1

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