Let R be the focus of the parabola y^2=20x, and the line y=mx+c intersect the parabola at …

Question

Let $R$ be the focus of the parabola $y^2=20x$, and the line $y=mx+c$ intersect the parabola at two points $P$ and $Q$. Let $G(10,10)$ be the centroid of triangle $PQR$. If $c-m=6$, then $(PQ)^2$ is:

Accepted Answer

Substituting $x=(y-c)/m$ into $y^2=20x$: $my^2-20y+20c=0$, so $y_1+y_2=20/m$.

Focus $R=(5,0)$. Centroid $y$-condition: $\dfrac{20/m+0}3=10\Rightarrow m=\dfrac23$. Then $c=6+\dfrac23=\dfrac{20}3$.

$$y^2-30y+200=0\ \Longrightarrow\ y=10,20\ \Longrightarrow\ P=(5,10),\ Q=(20,20).$$

$(PQ)^2=(20-5)^2+(20-10)^2=225+100=325.$

Solutions · 1