Find the equation of the hyperbola satisfying the give conditions: Foci $(\pm 5,\,0),$ the transverse axis is of length $8$
Foci $(\pm 5,\,0),$ the transverse axis 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 $(\pm 5,\,0)$, $c=5$
since the length of the transverse axis is $8,2 a=8 \Rightarrow a=4$
We know that $a^{2}+b^{2}=c^{2}$
$\therefore 4^{2}+b^{2}=52$
$\Rightarrow b^{2}=25-16=9$
Thus, the equation of the hyperbola is $\frac{x^{2}}{16}-\frac{y^{2}}{9}=1$
Find the equation of the hyperbola satisfying the give conditions: Foci $(\pm 4,\,0),$ the latus rectum is of length $12$
The length of the latus rectum and directrices of a hyperbola with eccentricity e are 9 and $\mathrm{x}= \pm \frac{4}{\sqrt{3}}$, respectively. Let the line $y-\sqrt{3} \mathrm{x}+\sqrt{3}=0$ touch this hyperbola at $\left(\mathrm{x}_0, \mathrm{y}_0\right)$. If $\mathrm{m}$ is the product of the focal distances of the point $\left(\mathrm{x}_0, \mathrm{y}_0\right)$, then $4 \mathrm{e}^2+\mathrm{m}$ is equal to ...........
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