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MHT CET · Maths · Hyperbola

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

  1. A 0
  2. B 1
  3. C 2
  4. D 3
Verified Solution

Answer & Solution

Correct Answer

(B) 1

Step-by-step Solution

Detailed explanation

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, \(\frac{1}{e_{1}^{2}}+\frac{1}{e_{2}^{2}}=\frac{9}{25}+\frac{16}{25}=1\)