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If $e_{1}$ and $e_{2}$ are the eccentricities of a hyperbola $3 x^{2}-3 y^{2}=25$ and its conjugate, then
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Verified Answer
The correct answer is:
$e_{1}^{2}+e_{2}^{2}=4$
We have, equation of hyperbola
$$
\begin{gathered}
3 x^{2}-2 j^{2}=25 \\
\Rightarrow \quad \frac{x^{2}}{\left(\frac{5}{\sqrt{3}}\right)^{2}}-\frac{y^{2}}{\left(\frac{5}{\sqrt{2}}\right)^{2}}=1
\end{gathered}
$$
Equation of conjugate hyperbola is
$$
\frac{y^{2}}{\left(\frac{5}{\sqrt{2}}\right)^{2}}-\frac{x^{2}}{\left(\frac{5}{\sqrt{3}}\right)^{2}}=1
$$
Now, eccentricity of hyperbola,
$$
\begin{aligned}
e_{1} &=\sqrt{1+\frac{\left(\frac{5}{\sqrt{2}}\right)^{2}}{\left(\frac{5}{\sqrt{3}}\right)^{2}}} \\
&=\sqrt{1+\frac{3}{2}}=\sqrt{\frac{5}{2}}
\end{aligned}
$$
and eccentricity of conjugate hyperbola,
$$
e_{2}=\sqrt{1+\frac{\left(\frac{5}{\sqrt{3}}\right)^{2}}{\left(\frac{5}{\sqrt{2}}\right)^{2}}}=\sqrt{1+\frac{2}{3}}=\sqrt{\frac{5}{3}}
$$
Now, $e_{1}^{2}+e_{2}^{2}=\frac{5}{2}+\frac{5}{3}=\frac{25}{6} \simeq 4$
$$
\begin{gathered}
3 x^{2}-2 j^{2}=25 \\
\Rightarrow \quad \frac{x^{2}}{\left(\frac{5}{\sqrt{3}}\right)^{2}}-\frac{y^{2}}{\left(\frac{5}{\sqrt{2}}\right)^{2}}=1
\end{gathered}
$$
Equation of conjugate hyperbola is
$$
\frac{y^{2}}{\left(\frac{5}{\sqrt{2}}\right)^{2}}-\frac{x^{2}}{\left(\frac{5}{\sqrt{3}}\right)^{2}}=1
$$
Now, eccentricity of hyperbola,
$$
\begin{aligned}
e_{1} &=\sqrt{1+\frac{\left(\frac{5}{\sqrt{2}}\right)^{2}}{\left(\frac{5}{\sqrt{3}}\right)^{2}}} \\
&=\sqrt{1+\frac{3}{2}}=\sqrt{\frac{5}{2}}
\end{aligned}
$$
and eccentricity of conjugate hyperbola,
$$
e_{2}=\sqrt{1+\frac{\left(\frac{5}{\sqrt{3}}\right)^{2}}{\left(\frac{5}{\sqrt{2}}\right)^{2}}}=\sqrt{1+\frac{2}{3}}=\sqrt{\frac{5}{3}}
$$
Now, $e_{1}^{2}+e_{2}^{2}=\frac{5}{2}+\frac{5}{3}=\frac{25}{6} \simeq 4$
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