Conic Sections Question 456
Question: If the foci of the ellipse $ \frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1 $ and the hyperbola $ \frac{x^{2}}{144}-\frac{y^{2}}{81}=\frac{1}{25} $ coincide, then the value of $ b^{2} $ is
[MNR 1992; UPSEAT 2001; AIEEE 2003; Karnataka CET 2004; Kerala (Engg.) 2005]
Options:
A) 1
B) 5
C) 7
D) 9
Show Answer
Answer:
Correct Answer: C
Solution:
Hyperbola is $ \frac{x^{2}}{144}-\frac{y^{2}}{81}=\frac{1}{25} $
$ a=\sqrt{\frac{144}{25}},b=\sqrt{\frac{81}{25}},e_1=\sqrt{1+\frac{81}{144}}=\sqrt{\frac{225}{144}}=\frac{15}{12}=\frac{5}{4} $
Therefore, foci $ =(ae_1,0)=( \frac{12}{5}.\frac{5}{4},0 )=(3,0) $
Therefore, focus of ellipse $ =(4e,0) $ i.e. $ (3,0) $
therefore $ x^{2}+ $
Hence $ b^{2}=16( 1-\frac{9}{16} )=7 $ .
BETA
