Question

The distance of the point $(6,-2\sqrt{2})$ from the common tangent $y=mx+c,m>0$, of the curves $x=2y^2$ and $x=1+y^2$ is :

• A. $5\sqrt{3}$
• B. $\frac{14}{3}$
• C. $\frac{1}{3}$
• D. 5

Answer 1

Answer: (D)

For
$y^2=\frac{x}{2},T:y=mx+\frac{1}{8m}$
For tangent to $y^2+1=x$
$\Rightarrow {\left(mx+\frac{1}{8m}\right)}^2+1=x$
$D=0\Rightarrow m=\frac{1}{2\sqrt{2}}$
$\therefore T:x-2\sqrt{2}y+1=0$
$d=\left|\frac{6+8+1}{\sqrt{9}}\right|=5$