Conic Sections Question 339
Question: An ellipse passes through the point (-3, 1) and its eccentricity is $ \sqrt{\frac{2}{5}} $ . The equation of the ellipse is
Options:
A) $ 3x^{2}+5y^{2}=32 $
B) $ 3x^{2}+5y^{2}=25 $
C) $ 3x^{2}+y^{2}=4 $
D) $ 3x^{2}+y^{2}=9 $
Show Answer
Answer:
Correct Answer: A
Solution:
Let the equation of ellipse be $ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 $
$ \because $ It passes through (- 3, 1) So $ \frac{9}{a^{2}}+\frac{1}{b^{2}}=1\Rightarrow 9+\frac{a^{2}}{b^{2}}=a^{2} $ ……(i) Given eccentricity is $ \sqrt{2/5} $
So $ \frac{2}{5}=1-\frac{b^{2}}{a^{2}}\Rightarrow \frac{b^{2}}{a^{2}}=\frac{3}{5} $ ……(ii) From equation (i) and (ii), $ a^{2}=\frac{32}{3},b^{2}=\frac{32}{5} $
Hence required equation of ellipse is $ 3x^{2}+5y^{2}=32 $ .
BETA
