Problem

The parabola $C$ has equation $y^2 = \dfrac{1}{8}x$

The point $P\left(\dfrac{t^2}{8}, \dfrac{t}{4}\right)$, where $t \neq 0$, lies on $C$.

(a) Use calculus to show that the tangent to $C$ at $P$ has equation

$8yt - 8x = t^2$

Given that

• the tangent to $C$ at $P$ meets the $y$-axis at the point $A$
• the line $l_1$ is the perpendicular bisector of the line segment $OA$ where $O$ is the origin
• the line $l_2$ is the perpendicular bisector of the line segment $AP$
• $l_1$ and $l_2$ intersect at the point $Q$

(b) show that, as $t$ varies, the coordinates of $Q$ satisfy the equation

$y^2 = \alpha x + \beta$

where $\alpha$ and $\beta$ are constants to be determined.