Find the equation of the hyperbola where foci are $(0,\,±12)$ and the length of the latus rectum is $36.$
Find the equation of the hyperbola where foci are $(0,\,±12)$ and the length of the latus rectum is $36.$
since foci are $(0,\,±12)$ , it follows that $c=12$
Length of the latus rectum $=\frac{2 b^{2}}{a}=36$ or $b^{2}=18 a$
Therefore $c^{2}=a^{2}+b^{2}$; gives
$144=a^{2}+18 a$
i.e. $a^{2}+18 a-144=0$
So $a=-24,6$
since $a$ cannot be negative, we take $a=6$ and so $b^{2}=108$.
Therefore, the equation of the required hyperbola is $\frac{y^{2}}{36}-\frac{x^{2}}{108}=1,$ i.e., $3 y^{2}-x^{2}=108$
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