JEE MAIN - Mathematics (2023 - 8th April Morning Shift - No. 8)
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 :
Let the point $$G(10,10)$$ be the centroid of the triangle $$P Q R$$. If $$c-m=6$$, then $$(P Q)^{2}$$ is :
317
325
346
296
Explanation
$$
y^2=20 x, y=m x+\mathrm{c}
$$
Put value of $x$
$$ \begin{aligned} & y^2=20\left(\frac{y-c}{m}\right) \\\\ & \Rightarrow y^2-\frac{20}{m} y+\frac{20}{m} c=0 .......(i) \end{aligned} $$
Since, centroid $=(10,10)$
$$ \begin{aligned} & \text { So, } \frac{y_1+y_2+0}{3}=10 \\\\ & \Rightarrow y_1+y_2=30 \end{aligned} $$
From (1),
$$ \text { Sum of roots }=\frac{20}{m}=30 \Rightarrow m=\frac{2}{3} $$
Also, $c-m=6 \Rightarrow c=6+\frac{2}{3}=\frac{20}{3}$
Now, the equation is :
$$ \begin{aligned} & y^2-\frac{20}{2} \times 3 y+\frac{20}{2} \times 3 \times \frac{20}{3}=0 \\\\ & \Rightarrow y^2-30 y+200=0 \\\\ & \Rightarrow y^2-20 y-10 y+200=0 \\\\ & \Rightarrow(y-20)(y-10)=0 \\\\ & \Rightarrow y=10,20 \Rightarrow x=5, x=20 \\\\ & \therefore P \equiv(5,10), Q \equiv(20,20) \\\\ & \text { So, }(P Q)^2=(20-5)^2+(20-10)^2 \\\\ & =225+100=325 \end{aligned} $$
Put value of $x$
$$ \begin{aligned} & y^2=20\left(\frac{y-c}{m}\right) \\\\ & \Rightarrow y^2-\frac{20}{m} y+\frac{20}{m} c=0 .......(i) \end{aligned} $$
Since, centroid $=(10,10)$
$$ \begin{aligned} & \text { So, } \frac{y_1+y_2+0}{3}=10 \\\\ & \Rightarrow y_1+y_2=30 \end{aligned} $$
From (1),
$$ \text { Sum of roots }=\frac{20}{m}=30 \Rightarrow m=\frac{2}{3} $$
Also, $c-m=6 \Rightarrow c=6+\frac{2}{3}=\frac{20}{3}$
Now, the equation is :
$$ \begin{aligned} & y^2-\frac{20}{2} \times 3 y+\frac{20}{2} \times 3 \times \frac{20}{3}=0 \\\\ & \Rightarrow y^2-30 y+200=0 \\\\ & \Rightarrow y^2-20 y-10 y+200=0 \\\\ & \Rightarrow(y-20)(y-10)=0 \\\\ & \Rightarrow y=10,20 \Rightarrow x=5, x=20 \\\\ & \therefore P \equiv(5,10), Q \equiv(20,20) \\\\ & \text { So, }(P Q)^2=(20-5)^2+(20-10)^2 \\\\ & =225+100=325 \end{aligned} $$
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