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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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The equation of the ellipse (referred to its axis as the axis of x and y respectively) which passes through the point $(-3,1)$ and had eccentricity $\sqrt {\large\frac{ 2}{5}}$

$\begin{array}{1 1}(A)\;3x^2+6y^2=33 \\(B)\;5x^2+3y^2=48 \\(C)\;3x^2+5y^2=32 \\(D)\;none\;of\;these \end{array}$

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1 Answer

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Let the equation of the required ellipse be $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1$
As it passses through $(-3,1)$ we get,
$\large\frac{9}{a^2}+\frac{1}{b^2}$$=2$
=> $9b^2+a^2=a^2b^2$
=> $9a^2(1-e^2)+a^2=a^2.a^2(1-e^2)$
=> $9a^2(1- \large\frac{2}{5})$$+a^2=a^4(1- \large\frac{2}{5})$
=> $a^2=\large\frac{32}{5}$
Now $b^2=a^2(1-e^2)$
=> $b^2=\large\frac{32}{3} (1-\large\frac{2}{5})$
$\qquad= \large\frac{32}{5}$
Hence the equation of required ellipse is
$\large\frac{x^2}{\Large\frac{32}{3}}+\frac{y^2}{\Large\frac{32}{5}}$$=1$
(or) $3x^2+5y^2=32$
Hence C is the correct answer.
answered Apr 10, 2014 by meena.p
 

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