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Q.
The equation of the hyperbola in the standard form (with transverse axis along the $x$-axis) having the length of the latus rectum = $9$ units and eccentricity = $5/4$ is
Conic Sections
Solution:
Let equation of hyperbola be $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$
$\therefore e=\sqrt{1+\frac{b^{2}}{a^{2}}}\Rightarrow \left(\frac{5}{4}\right)^{2}$
$=\left(1+\frac{b^{2}}{a^{2}}\right)\left[\because e=\frac{5}{4}\right]$
$\Rightarrow \frac{b^{2}}{a^{2}}=\frac{9}{16}$
$\Rightarrow \frac{b}{a}=\frac{3}{4}$
We have, $\frac{2b^{2}}{a}=9$
$\Rightarrow 2\times\frac{3}{4}b=9$
$\Rightarrow b = 6$ and $a = 8$
$\therefore $ Equation of hyperbola is $\frac{x^{2}}{64}-\frac{y^{2}}{36}=1$