The distance between the tangents to the hyperbola $\frac{x^2}{20}-\frac{3y^2}{4}=1$ which are parallel to the line $x+3y=7$ is

Given

Hyperbola $\frac{x^2}{20}-\frac{3y^2}{4}=1$ and tangents are parallel to the line $x+3y=7$ 

So we have $a^2=20, b^2=\frac{4}{3}, m=\frac{-1}{3} ...\left(\mathrm{i}\right)$

We know that equation of tangent of a hyperbola in slope form is given by $y=mx\pm \sqrt{a^2{m}^2-b^2}$

Using the Equation $\left(\mathrm{i}\right)$

$y=\frac{-1}{3}x\pm \frac{2\sqrt{2}}{3}\Rightarrow 3y=-x\pm 2\sqrt{2}$

So tangents are $3y+x=2\sqrt{2}, 3y+x=-2\sqrt{2}$

Now distance between parallel lines 

$=\left|\frac{4\sqrt{2}}{\sqrt{9+1}}\right|=\frac{4\sqrt{2}}{\sqrt{10}}=\frac{4}{\sqrt{5}}$