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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:
$\mathrm{e}_{1}^{2}+\mathrm{e}_{2}^{2}=4$
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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