Conic Sections Question 224
Question: The latus rectum of the hyperbola $ 9x^{2}-16y^{2}-18x-32y-151=0 $ is
[MP PET 1996]
Options:
A) $ \frac{9}{4} $
B) 9
C) $ \frac{3}{2} $
D) $ \frac{9}{2} $
Show Answer
Answer:
Correct Answer: D
Solution:
$ 9x^{2}-18x+9-16y^{2}-32y-16=144 $
$ \Rightarrow \frac{{{(x-1)}^{2}}}{16}-\frac{{{(y+1)}^{2}}}{9}=1 $
therefore Latus rectum $ =\frac{2b^{2}}{a}=\frac{2\times 9}{4}=\frac{9}{2} $ .
BETA
