Conic Sections Question 396
Question: If the eccentricities of the hyperbolas $ \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 $ and $ \frac{y^{2}}{b^{2}}-\frac{x^{2}}{a^{2}}=1 $ be e and $ e_1 $ , then $ \frac{1}{e^{2}}+\frac{1}{e_1^{2}}= $
[MNR 1984; MP PET 1995; DCE 2000]
Options:
A) 1
B) 2
C) 3
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ e=\sqrt{1+\frac{b^{2}}{a^{2}}} $
therefore $ e^{2}=\frac{a^{2}+b^{2}}{a^{2}} $
$ e_1=\sqrt{1+\frac{a^{2}}{b^{2}}} $
therefore $ e_1^{2}=\frac{b^{2}+a^{2}}{b^{2}} $
therefore $ \frac{1}{e_1^{2}}+\frac{1}{e^{2}}=1 $ .
BETA
