Question

A circle touches the parabola $y^2=4x$ at $M(1,2)$ and also touches its directrix. The $y$-coordinate of the point of contact of the circle and the directrix is

• A. $\sqrt{2}$
  
• B. $2$
  
• C. $2\sqrt{2}$
  
• D. $4$

Answer 1

The correct option is C $2\sqrt{2}$


$y^2=4x$
Here, $a=1$
Hence, equation of directrix is $T_1:x=-1$
Slope of tangent at $(1,2)$ to $y^2=4x$ is $m=1$
So, equation of tangent at $(1,2)$ is
$T_2:y-2=1(x-1)$
or $T_2:y=x+1$
Intersection of the tangents $T_1$ and $T_2$ is $(-1,0)$
So, $T_1$ and $T_2$ are two tangents from point $(-1,0)$
We know that tangents drawn from an external point to the circle are equal in length.
Distance between $(-1,0)$ and $(1,2)$ is $\sqrt{2^2+2^2}=2\sqrt{2}$
Hence, the coordinates of the point of contact of the circle and the directrix are $(-1,2\sqrt{2})$