AP EAMCET · Maths · Hyperbola
If \(e_1, e_2\) are respectively the eccentricities of the curves \(9 x^2-16 y^2-144=0\) and \(9 x^2-16 y^2+144=0\), then \(\frac{e_1^2 e_2^2}{e_1^2+e_2^2}=\)
- A \(\sqrt{2}\)
- B 1
- C \(\sqrt{3}\)
- D 2
Answer & Solution
Correct Answer
(B) 1
Step-by-step Solution
Detailed explanation
(i) is a hyperbola and (ii) is a conjugate hyperbola, We have, \[ \begin{array}{ll} \therefore & \frac{1}{e_1^2}+\frac{1}{e_2^2}=1 \\ \therefore & \frac{e_1^2+e_2^2}{e_1^2 e_2^2}=1 \Rightarrow \frac{e_1^2 e_2^2}{e_1^2+e_2^2}=1 \end{array} \] Hence, option (b) is correct.
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