Let the normal at a point on the parabola y^2=6x pass through the point (5,-8).

Question

Let the normal at a point on the parabola $y^2=6x$ pass through the point $(5,-8)$. If the tangent at $P$ to the parabola intersects its directrix at the point $Q$, then the ordinate of $Q$ is:

Accepted Answer

Parabola: $a=\tfrac32$, $P=(\tfrac32t^2,3t)$. Normal: $y=-tx+3t+\tfrac32t^3$. Through $(5,-8)$:
$$3t^3-4t+16=0\ \Longrightarrow\ (t+2)(3t^2-6t+8)=0\ \Longrightarrow\ t=-2$$
(the quadratic factor has no real roots). So $P=(6,-6)$.

Tangent at $P$: $ty=x+at^2\Rightarrow-2y=x+6\Rightarrow x=-2y-6$. Directrix: $x=-\tfrac32$:
$$-\frac32=-2y-6\ \Longrightarrow\ y=-\frac94.$$

Solutions · 1