Conic Sections Question 312
Question: The eccentricity of the conic $ 4x^{2}+16y^{2}-24x-3y=1 $ is
[MP PET 2004]
Options:
A) $ \frac{\sqrt{3}}{2} $
B) $ \frac{1}{2} $
C) $ \frac{\sqrt{3}}{4} $
D) $ \sqrt{3} $
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Answer:
Correct Answer: A
Solution:
Given equation of conic is $ 4x^{2}+16y^{2}-24x-3y=1 $
$ \Rightarrow $ $ {{(2x-6)}^{2}}+{{(4y-4)}^{2}}=53 $
$ \Rightarrow $ $ 4{{(x-3)}^{2}}+16{{(y-1)}^{2}}=53 $
$ \Rightarrow $ $ \frac{{{(x-3)}^{2}}}{53/4}+\frac{{{(y-1)}^{2}}}{53/16}=1 $
$ \therefore $ $ e=\sqrt{1-\frac{b^{2}}{a^{2}}}=\sqrt{1-\frac{53/16}{53/4}}=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2} $ .
BETA
