Let the eccentricity of the hyperbola $H:\frac{x^2}{a^2}-\frac{y^2}{{ b}^2}=1$ be $\sqrt{\frac{5}{2}}$ and length of its latus rectum be $6\sqrt{2}$. If $y=2x+c$ is a tangent to the hyperbola $H$, then the value of $c^2$ is equal to

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 $c^2=20$