Conic Sections Question 318
Question: The vertices of a hyperbola are at (0, 0) and (10, 0) and one of its foci is at (18, 0). The equation of the hyperbola is
Options:
A) $ \frac{x^{2}}{25}-\frac{y^{2}}{144}=1 $
B) $ \frac{{{(x-5)}^{2}}}{25}-\frac{y^{2}}{144}=1 $
C) $ \frac{x^{2}}{25}-\frac{{{(y-5)}^{2}}}{144}=1 $
D) $ \frac{{{(x-5)}^{2}}}{25}-\frac{{{(y-5)}^{2}}}{144}=1 $
Show Answer
Answer:
Correct Answer: B
Solution:
$ 2a=10 $ , $ a=5 $
$ ae-a=8 $ or $ e=1+\frac{8}{5}=\frac{13}{5} $ $ b=5\sqrt{\frac{13^{2}}{5^{2}}-1}=5\times \frac{12}{5}=12 $ and centre of hyperbola $ \equiv (5,0) $
$ \frac{{{(x-5)}^{2}}}{5^{2}}-\frac{{{(y-0)}^{2}}}{12^{2}}=1 $ .
BETA
