Equation of hyperbola whose foci are foci of ellipse frac{{{x^2}}}{{25}} + frac{{{y^2}}}{9} = 1 and

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10-2. Parabola, Ellipse, Hyperbola

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The equation of the hyperbola whose foci are the foci of the ellipse $\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{9} = 1$ and the eccentricity is $2$, is

A

$\frac{{{x^2}}}{4} + \frac{{{y^2}}}{{12}} = 1$

B

$\frac{{{x^2}}}{4} - \frac{{{y^2}}}{{12}} = 1$

C

$\frac{{{x^2}}}{{12}} + \frac{{{y^2}}}{4} = 1$

D

$\frac{{{x^2}}}{{12}} - \frac{{{y^2}}}{4} = 1$

Solution

(b) Here for given ellipse $a = 5,\;b = 3,\;{b^2} = {a^2}(1 – {e^2})$

==> $e = \frac{4}{5}$

Therefore, focus is $(-4, 0), (4, 0).$

Given eccentricity of hyperbola = $2$

$a = \frac{{ae}}{e} = \frac{4}{2} = 2$ and $b = 2\sqrt {(4 – 1)} = 2\sqrt 3 $

Hence hyperbola is $\frac{{{x^2}}}{4} – \frac{{{y^2}}}{{12}} = 1$.

Standard 11

Mathematics

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