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

Question

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

Accepted Answer

${y}^{2}=20x,y=mx+c$

Put value of $x$

$\begin{gathered} {y}^{2}=20(\frac{y-c}{m}) \ \Rightarrow {y}^{2}-\frac{20}{m}y+\frac{20}{m}c=0.......(i) \end{gathered}$

Since, centroid $=(10,10)$

$\begin{gathered} \frac{{y}_{1}+{y}_{2}+0}{3}=10 \ \Rightarrow {y}_{1}+{y}_{2}=30 \end{gathered}$

From (1),

$=\frac{20}{m}=30\Rightarrow m=\frac{2}{3}$

Also, $c-m=6\Rightarrow c=6+\frac{2}{3}=\frac{20}{3}$

Now, the equation is :

$\begin{gathered} {y}^{2}-\frac{20}{2}	imes 3y+\frac{20}{2}	imes 3	imes \frac{20}{3}=0 \ \Rightarrow {y}^{2}-30y+200=0 \ \Rightarrow {y}^{2}-20y-10y+200=0 \ \Rightarrow (y-20)(y-10)=0 \ \Rightarrow y=10,20\Rightarrow x=5,x=20 \ 	herefore P\equiv (5,10),Q\equiv (20,20) \ (PQ{)}^{2}=(20-5{)}^{2}+(20-10{)}^{2} \ =225+100=325 \end{gathered}$

Solutions · 1