NIMCET 2011 Question Paper with solutions

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Question : 19 of 120

Marks: +1, -0

The eccentricity of ellipse

9 x^{2}+5 y^{2}-30 y=0

is

$1 / 3$
$2 / 3$
$3 / 4$
$1 / 4$

Solution:

Given.

9 x^{2}+5 y^{2}-30 y=0

⇒ 9 x^{2}+5(y^{2}-6 y)=0

⇒ 9 x^{2}+5(y-3)^{2}=45

⇒ {x^{2}}/{5}+ {(y-3)^{2}}/{9}=1

Here,

a^{2}=5, b^{2}=9

and

b>a

∴ e=√{1-\;{a^{2}}/{b^{2}}}=√{1- {5}/{9}}= {2}/{3}

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