Conic Sections Question 230
Question: The hyperbola $ \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 $ passes through the point $ (3\sqrt{5},1) $ and the length of its laths rectum is $ \frac{4}{3} $ units. The length of the conjugate axis is
Options:
A) 2 units
B) 3 units
C) 4 units
D) 5 units
Show Answer
Answer:
Correct Answer: C
Solution:
[c] $ \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 $ Hyperbola passes through $ (3\sqrt{5},1) $
$ \therefore \frac{{{(3\sqrt{5})}^{2}}}{a^{2}}-\frac{1}{b^{2}}=1 $ $ \frac{45}{a^{2}}-\frac{1}{b^{2}}=1 $
…… (i) Now length of latus rectum $ =\frac{2b^{2}}{a} $
$ \Rightarrow \frac{4}{3}=\frac{2b^{2}}{a} $
$ \Rightarrow \frac{2}{3}=\frac{b^{2}}{a}\Rightarrow a=\frac{3b^{2}}{2} $
…… (ii) Putting the value of -a- for equation (ii) in equation (i),
$ \Rightarrow \frac{45\times 4}{9b^{2}}-\frac{1}{b^{2}}=1\Rightarrow \frac{20}{b^{4}}-\frac{1}{b^{2}}=1 $ $ 20-b^{2}=b^{4} $ $ b^{4}+b^{2}-20=0 $ $ b^{4}+5b^{2}-4b^{2}-20=0 $ $ b^{2}(b^{2}+5)-4(b^{2}+5)=0 $ $ (b^{2}-4)(b^{2}+5)=0 $ $ b^{2}=4,b^{2}=-5 $
$ \therefore b^{2}=4\Rightarrow b=2 $ Now length of conjugate axis $ =2b=2(2)=4 $
BETA
