Answer:
Option B
Explanation:
Equation of tangent to hyperbola $\frac{x^{2}}{20}- \frac{3y^{2}}{4}$=1
parallel to x+3y=7 is
$y=-\frac{1}{3}x\pm\sqrt{\frac{20}{9}-\frac{4}{3}}$
$3y=-x\pm2\sqrt{2}$
$x+3y=2\sqrt{2}$
and $x+3y=-2\sqrt{2}$
Distance between parallel tangent is
$d=\frac{2\sqrt{2}+2\sqrt{2}}{\sqrt{1+3^{2}}}=\frac{4\sqrt{2}}{\sqrt{10}}=\frac{4}{\sqrt{5}}$
