VJC P1 Q6

A curve has equation $y^2=4x$ and a line $l$ has equation $2x-y+1=0$ as shown above.

$B(b, 2\sqrt{b})$ is a fixed point on C and A is an arbitrary point on $l$. State the geometrical relationship between the line segment AB and $l$ is the distance from B to A is the least.

Taking the coordinates of A as $(a, 2a+1)$, find an equation relating $a$ and $b$ for which AB is the least.

Deduce that when AB is the least, $(AB)^2 = m (2b - 2\sqrt{b} +1)^2$ where $m$ is a constant to be found. Hence or otherwise, find the coordinates of the point on C that is nearest on $l$.