Let y^2 = 12x be the parabola with its vertex at O.

Question

Let $y^2 = 12x$ be the parabola with its vertex at $O$. Let $P$ be a point on the parabola and $A$ be a point on the $x$-axis such that $\angle OPA = 90^\circ$. Then the locus of the centroid of such triangles $OPA$ is:

Accepted Answer

Let $a=3$ (since $4a=12$), $P=(3t^2,6t)$, $A=(x,0)$. Slope $OP=\dfrac{2}{t}$, slope $PA=\dfrac{6t}{3t^2-x}$.

$$\frac{2}{t}\cdot\frac{6t}{3t^2-x}=-1 \;\Rightarrow\; \frac{12}{3t^2-x}=-1 \;\Rightarrow\; x=3t^2+12.$$

Centroid $(h,k)$: $h=\dfrac{0+3t^2+x}{3}=2t^2+4$, $k=\dfrac{6t}{3}=2t$.

$$t=\frac k2 \;\Rightarrow\; h=2\left(\frac{k}{2}\right)^2+4=\frac{k^2}{2}+4 \;\Rightarrow\; k^2-2h+8=0,$$

i.e. $y^2-2x+8=0$.

Solutions · 1