Search any question & find its solution
Question:
Answered & Verified by Expert
If $e_{1}$ and $e_{2}$ represent the eccentricity of the curves $16 x^{2}-9 y^{2}=144$ and $9 x^{2}-16 y^{2}=144$
respectively. Then, $\frac{1}{e_{1}^{2}}+\frac{1}{e_{2}^{2}}$ is equal to
Options:
respectively. Then, $\frac{1}{e_{1}^{2}}+\frac{1}{e_{2}^{2}}$ is equal to
Solution:
2582 Upvotes
Verified Answer
The correct answer is:
1
Given curves are $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$ and $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$
$\therefore$
$e_{1}=\sqrt{1+\frac{16}{9}}=\frac{5}{3}$
and $e_{2}=\sqrt{1+\frac{9}{16}}=\frac{5}{4}$
Now, $\frac{1}{e_{1}^{2}}+\frac{1}{e_{2}^{2}}=\frac{9}{25}+\frac{16}{25}=1$
$\therefore$
$e_{1}=\sqrt{1+\frac{16}{9}}=\frac{5}{3}$
and $e_{2}=\sqrt{1+\frac{9}{16}}=\frac{5}{4}$
Now, $\frac{1}{e_{1}^{2}}+\frac{1}{e_{2}^{2}}=\frac{9}{25}+\frac{16}{25}=1$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.