Question

The distance of the point $(6,-2 \sqrt{2})$ from the common tangent $y=m x+c, m>0$, of the curves $x=2 y^2$ 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
$y ^2=\frac{ x }{2}, T : y = mx +\frac{1}{8 m }$
For tangent to $y^2+1=x$
$\Rightarrow\left( mx +\frac{1}{8 m }\right)^2+1=x $
$D =0 \Rightarrow m =\frac{1}{2 \sqrt{2}} $
$\therefore T : x -2 \sqrt{2} y +1=0 $
$d =\left|\frac{6+8+1}{\sqrt{9}}\right|=5$