The difference between the focal distances of any point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$…
The difference between the focal distances of any point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is 6 . If $(\sqrt{13}, k)$ is an end point of a latus rectum of this hyperbola, then $\mathrm{k}=$
$\pm \frac{9}{2}$
$\pm \frac{8}{3}$
$\pm {9}$
$\pm \frac{4}{3}$
Solution
We have difference between foci $=2 a$
$
\therefore 2 a=6 \Rightarrow a=3 \text {, }
$
$\because$ End of focal chord is $\left(c, \pm \frac{b^2}{a}\right) \Rightarrow c=\sqrt{13}$
$
\because c^2=a^2+b^2 \Rightarrow 13=9+b^2 \Rightarrow b^2=4
$
$
\therefore k= \pm \frac{b^2}{a}= \pm \frac{4}{3} \text {. }
$