If foci of an ellipse are ( pm √ 5 ,,0) and its eccentricity is frac{{√ 5 }}{3} , then equation

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10-2. Parabola, Ellipse, Hyperbola

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If the foci of an ellipse are $( \pm \sqrt 5 ,\,0)$ and its eccentricity is $\frac{{\sqrt 5 }}{3}$, then the equation of the ellipse is

A

$9{x^2} + 4{y^2} = 36$

B

$4{x^2} + 9{y^2} = 36$

C

$36{x^2} + 9{y^2} = 4$

D

$9{x^2} + 36{y^2} = 4$

Solution

(b) $\because \,ae = \pm \sqrt 5 $

==> $a = \pm \sqrt 5 \left( {\frac{3}{{\sqrt 5 }}} \right) = \pm 3$

==> ${a^2} = 9$

${b^2} = {a^2}(1 – {e^2}) = 9\left( {1 – \frac{5}{9}} \right) = 4$

Hence, equation of ellipse $\frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} = 1$

==> $4{x^2} + 9{y^2} = 36$.

Standard 11

Mathematics

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