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If the eccentricity of a hyperbola is $\frac{5}{3}$, then the eccentricity of its conjugate hyperbola is
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The correct answer is:
$\frac{5}{4}$
Eccentricity of given hyperbola $=\frac{5}{3}$
Let, eccentricity of its conjugate hyperbola $=e$
So, $\frac{1}{\left(\frac{5}{3}\right)^2}+\frac{1}{e^2}=1$
$\Rightarrow \quad \frac{9}{25}+\frac{1}{e^2}=1 \quad \Rightarrow \frac{1}{e^2}=1-\frac{9}{25}$
$\Rightarrow \quad \frac{1}{e^2}=\frac{16}{25} \Rightarrow e^2=\frac{25}{16}$
$$
\Rightarrow \quad e=\frac{5}{4}
$$
Hence, eccentricity of conjugate hyperbola $=\frac{5}{4}$
Let, eccentricity of its conjugate hyperbola $=e$
So, $\frac{1}{\left(\frac{5}{3}\right)^2}+\frac{1}{e^2}=1$
$\Rightarrow \quad \frac{9}{25}+\frac{1}{e^2}=1 \quad \Rightarrow \frac{1}{e^2}=1-\frac{9}{25}$
$\Rightarrow \quad \frac{1}{e^2}=\frac{16}{25} \Rightarrow e^2=\frac{25}{16}$
$$
\Rightarrow \quad e=\frac{5}{4}
$$
Hence, eccentricity of conjugate hyperbola $=\frac{5}{4}$
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