Consider the parabola with vertex (12,34) and directrix y=12.

Question

Consider the parabola with vertex $\left(\dfrac12,\dfrac34\right)$ and directrix $y=\dfrac12$. Let $P$ be the point where the parabola meets the line $x=-\dfrac12$. If the normal to the parabola at $P$ intersects the parabola again at the point $Q$, then $(PQ)^2$ is equal to

Accepted Answer

Distance vertex-directrix $=\dfrac14=A$: $\left(x-\dfrac12\right)^2=\left(y-\dfrac34\right)$.

At $x=-\dfrac12$: $1=y-\dfrac34\Rightarrow y=\dfrac74$; $P=\left(-\dfrac12,\dfrac74\right)$.

Tangent slope $=2(x-\tfrac12)=-2$ at $P$; normal slope $=\dfrac12$: $y=\dfrac x2+2$.

Substituting into the parabola: $\left(x-\dfrac12\right)^2=\dfrac x2+2-\dfrac34=\dfrac x2+\dfrac54\Rightarrow2x^2-3x-2=0\Rightarrow x=2\text{ or }-\dfrac12$.

$x=2\Rightarrow y=3$; $Q=(2,3)$.

$$(PQ)^2=\left(2+\frac12\right)^2+\left(3-\frac74\right)^2=\left(\frac52\right)^2+\left(\frac54\right)^2=\frac{25}4+\frac{25}{16}=\frac{125}{16}.$$

Solutions · 1