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 :

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Answer: B. 325 y^2=20 x, y=m x+\mathrm{c} Put value of x \begin{aligned} & y^2=20\left(\frac{y-c}{m}ight) \\ & \Rightarrow y^2-\frac{20}{m} y+\frac{20}{m} c=0 .......(i) \end{aligned} Since, centroid =(10,10) \begin{aligned} & 	ext { So, } \frac{y_1+y_2+0}{3}=10 \\ & \Rightarrow y_1+y_2=30 \end{aligned} From (1), 	ext { Sum of roots }=\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{aligned} & y^2-\frac{20}{2} 	imes 3 y+\frac{20}{2} 	imes 3 	imes \frac{20}{3}=0 \\ & \Rightarrow y^2-30 y+200=0 \\ & \Rightarrow y^2-20 y-10 y+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) \\ & 	ext { So, }(P Q)^2=(20-5)^2+(20-10)^2 \\ & =225+100=325 \end{aligned}

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 :

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