Let the eccentricity of the hyperbola H : x 2 a 2 - y 2   b 2 = 1 be 5 2 and length of its latus rectum…

Given equation of the hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

If $e$ is eccentricity of the hyperbola, then $e^2=1+\frac{b^2}{a^2}$

$1+\frac{b^2}{a^2}=\frac{5}{2}$$\Rightarrow b^2=\frac{3a^2}{2}$

Length of the latus-rectum is given by $\frac{2b^2}{a}$

i.e. $\frac{2b^2}{a}=6\sqrt{2}$

$\frac{2}{a}\times \frac{3a^2}{2}=6\sqrt{2}$

$\Rightarrow a=2\sqrt{2}$

and $b^2=\frac{3}{2}\times 8=12$

$\Rightarrow b=2\sqrt{3}$

If $y=mx+c$ is a tangent to the hyperbola, then $c^2=a^2{m}^2-b^2$

$\therefore c^2=4\times 8-12$

Hence $c^2=20$