Conic Sections Question 501
Question: The latus rectum of the hyperbola $ 9x^{2}-16y^{2}+72x-32y-16=0 $ is
[Pb. CET 2004]
Options:
A) $ \frac{9}{2} $
B) $ -\frac{9}{2} $
C) $ \frac{32}{3} $
D) $ -\frac{32}{3} $
Show Answer
Answer:
Correct Answer: A
Solution:
Given equation of hyperbola is, $ 9x^{2}-16y^{2}+72x-32y-16=0 $
therefore $ 9(x^{2}+8x)-16(y^{2}+2y)-16=0 $
therefore $ 9{{(x+4)}^{2}}-16{{(y+1)}^{2}}=144 $
therefore $ \frac{{{(x+4)}^{2}}}{16}-\frac{{{(y+1)}^{2}}}{9}=1 $ Therefore, latus rectum = $ \frac{2b^{2}}{a}=2\times \frac{9}{4}=\frac{9}{2} $ .
BETA
