Let P be a point on the parabola y^2=12x and N be the foot of the perpendicular drawn from…

Question

Let $P$ be a point on the parabola $y^2=12x$ and $N$ be the foot of the perpendicular drawn from $P$ on the axis of the parabola. A line is now drawn through the mid-point $M$ of $PN$, parallel to its axis, which meets the parabola at $Q$. If the $y$-intercept of the line $NQ$ is $\dfrac43$, then:

Accepted Answer

$P=(3t^2,6t)$ (using $a=3$), $N=(3t^2,0)$, $M=(3t^2,3t)$.

Line $y=3t$ meets the parabola again at $Q=\left(\dfrac{3t^2}4,3t\right)$.

Line $NQ$ has slope $\dfrac{3t-0}{3t^2/4-3t^2}=-\dfrac4{3t}$; at $x=0$: $y$-intercept $=4t=\dfrac43\Rightarrow t=\dfrac13$.

$$MQ=\left|3t^2-\frac{3t^2}4\right|=\frac{9t^2}4=\frac9{4\cdot9}=\frac14.$$

Solutions · 1