KCET · Maths · Hyperbola
If \(e_{1}\) and \(e_{2}\) are the eccentricities of a hyperbola \(3 x^{2}-3 y^{2}=25\) and its conjugate, then
- A \(e_{1}^{2}+e_{2}^{2}=2\)
- B \(\mathrm{e}_{1}^{2}+\mathrm{e}_{2}^{2}=4\)
- C \(e_{1}+e_{2}=4\)
- D \(e_{1}+e_{2}=\sqrt{2}\)
Answer & Solution
Correct Answer
(B) \(\mathrm{e}_{1}^{2}+\mathrm{e}_{2}^{2}=4\)
Step-by-step Solution
Detailed explanation
Given equation can be rewritten as
\[
x^{2}-y^{2}=\frac{25}{3}
\]
Here, \(\quad \mathrm{a}^{2}=1, \mathrm{~b}^{2}=1\) \(\therefore \quad \mathrm{e}_{1}=\sqrt{1+\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}}=\sqrt{1+1}=\sqrt{2}\)
The equation of conjugate hyperbola is
\[
\begin{gathered}
-\mathrm{x}^{2}+\mathrm{y}^{2}=\frac{25}{3} \\
\therefore \quad \mathrm{e}_{2}=\sqrt{1+\frac{\mathrm{a}^{2}}{\mathrm{~b}^{2}}}=\sqrt{1+1}=\sqrt{2} \\
\therefore \quad \mathrm{e}_{1}^{2}+\mathrm{e}_{2}^{2}=(\sqrt{2})^{2}+(\sqrt{2})^{2}=4
\end{gathered}
\]
\[
x^{2}-y^{2}=\frac{25}{3}
\]
Here, \(\quad \mathrm{a}^{2}=1, \mathrm{~b}^{2}=1\) \(\therefore \quad \mathrm{e}_{1}=\sqrt{1+\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}}=\sqrt{1+1}=\sqrt{2}\)
The equation of conjugate hyperbola is
\[
\begin{gathered}
-\mathrm{x}^{2}+\mathrm{y}^{2}=\frac{25}{3} \\
\therefore \quad \mathrm{e}_{2}=\sqrt{1+\frac{\mathrm{a}^{2}}{\mathrm{~b}^{2}}}=\sqrt{1+1}=\sqrt{2} \\
\therefore \quad \mathrm{e}_{1}^{2}+\mathrm{e}_{2}^{2}=(\sqrt{2})^{2}+(\sqrt{2})^{2}=4
\end{gathered}
\]
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