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Find the equation of the hyperbola satisfying the give conditions: Foci $(\pm 3 \sqrt{5},\,0),$ the latus rectum is of length $8$
$\frac{x^{2}}{25}-\frac{y^{2}}{20}=1$
$\frac{x^{2}}{25}-\frac{y^{2}}{20}=1$
$\frac{x^{2}}{25}-\frac{y^{2}}{20}=1$
$\frac{x^{2}}{25}-\frac{y^{2}}{20}=1$
Solution
Foci $(\pm 3 \sqrt{5},\, 0),$ the latus rectum is of length $8$.
Here, the foci are on the $x-$ axis.
Therefore, the equation of the hyperbola is of the form $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$
since the foci are $(\pm 3 \sqrt{5}, \,0)$, $c=\pm 3 \sqrt{5}$
Length of latus rectum $=8$
$\Rightarrow \frac{2 b^{2}}{a}=8$
$\Rightarrow b^{2}=4 a$
We know that $a^{2}+b^{2}=c^{2}$
$\therefore a^{2}+4 a=45$
$\Rightarrow a^{2}+4 a-45=0$
$\Rightarrow a^{2}+9 a-5 a-45=0$
$\Rightarrow(a+9)(a-5)=0$
$\Rightarrow a=-9,5$
since a is non-negative, $a=5$
$\therefore b^{2}=4 a=4 \times 5=20$
Thus, the equation of the hyperbola is $\frac{x^{2}}{25}-\frac{y^{2}}{20}=1$
