Conic Sections Question 101
Question: The eccentricity of an ellipse, with its centre at the origin, is $ \frac{1}{2} $ . If one of the directrices is $ x=4 $ , then the equation of the ellipse is
[AIEEE 2004]
Options:
A) $ 4x^{2}+3y^{2}=1 $
B) $ 3x^{2}+4y^{2}=12 $
C) $ 4x^{2}+3y^{2}=12 $
D) $ 3x^{2}+4y^{2}=1 $
Show Answer
Answer:
Correct Answer: B
Solution:
Since directrix is parallel to y-axis, hence axes of the ellipse are parallel to x-axis. Let the equation of the ellipse be $ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 $ , $ (a>b) $
$ e^{2}=1-\frac{b^{2}}{a^{2}}\Rightarrow \frac{b^{2}}{a^{2}}=1-e^{2}=1-\frac{1}{4}\Rightarrow \frac{b^{2}}{a^{2}}=\frac{3}{4} $ . Also, one of the directrices is $ x=4 $
$ \Rightarrow $ $ \frac{a}{e}=4\Rightarrow a=4e=4.\frac{1}{2}=2 $ ; $ b^{2}=\frac{3}{4}a^{2}=\frac{3}{4}.4=3 $
$ \therefore $ Required ellipse is $ \frac{x^{2}}{4}+\frac{y^{2}}{3}=1 $ or $ 3x^{2}+4y^{2}=12 $ .
BETA
