JEE MAIN - Mathematics (2011 - No. 16)
Equation of the ellipse whose axes of coordinates and which passes through the point $$(-3,1)$$ and has eccentricity $$\sqrt {{2 \over 5}} $$ is :
$$5{x^2} + 3{y^2} - 48 = 0$$
$$3{x^2} + 5{y^2} - 15 = 0$$
$$5{x^2} + 3{y^2} - 32 = 0$$
$$3{x^2} + 5{y^2} - 32 = 0$$
Explanation
Let the ellipse be $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$
It press through $$(-3, 1)$$ so $${9 \over {{a^2}}} + {1 \over {{b^2}}} = 1\,\,\,\,\,\,...\left( i \right)$$
Also, $${b^2} = {a^2}\left( {1 - 2/5} \right)$$
$$ \Rightarrow 5{b^2} = 3{a^2}\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$
Solving $$(i)$$ and $$(ii)$$ we get $${a^2} = {{32} \over 3},{b^2} = {{32} \over 5}$$
So, the equation of the ellipse is $$3{x^2} + 5{y^2} = 32$$
It press through $$(-3, 1)$$ so $${9 \over {{a^2}}} + {1 \over {{b^2}}} = 1\,\,\,\,\,\,...\left( i \right)$$
Also, $${b^2} = {a^2}\left( {1 - 2/5} \right)$$
$$ \Rightarrow 5{b^2} = 3{a^2}\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$
Solving $$(i)$$ and $$(ii)$$ we get $${a^2} = {{32} \over 3},{b^2} = {{32} \over 5}$$
So, the equation of the ellipse is $$3{x^2} + 5{y^2} = 32$$
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