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A hyperbola having its centre at the origin is passing through the point $(5,2)$ and has transverse axis of length 8 along the $\mathrm{X}$-axis. Then the eccentricity of its conjugate hyperbola is
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The correct answer is:
$\frac{\sqrt{13}}{2}$
Let equation of hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
Given, $2 \mathrm{a}=8 \Rightarrow \mathrm{a}=4$
It passes through $(5,2)$
Hence, on solving we get, $\mathrm{b}=\frac{8}{3}$
Now, eccentricity of its conjugate hyperbola is
$$
=\sqrt{\frac{\mathrm{a}^2+\mathrm{b}^2}{\mathrm{~b}^2}}=\frac{\sqrt{13}}{2}
$$
Given, $2 \mathrm{a}=8 \Rightarrow \mathrm{a}=4$
It passes through $(5,2)$
Hence, on solving we get, $\mathrm{b}=\frac{8}{3}$
Now, eccentricity of its conjugate hyperbola is
$$
=\sqrt{\frac{\mathrm{a}^2+\mathrm{b}^2}{\mathrm{~b}^2}}=\frac{\sqrt{13}}{2}
$$
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