JEE Mains · Maths · STD 11 - 10.2 parabola,ellipse,hyperbola
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 \(( PQ )^2\) is
- A \(325\)
- B \(317\)
- C \(296\)
- D \(346\)
Answer & Solution
Correct Answer
(A) \(325\)
Step-by-step Solution
Detailed explanation
\(y^2=20 x, y=m x+c\) \(y^2=20\left(\frac{y-c}{m}\right)\) \(y^2-\frac{20 y}{m}+\frac{20 c}{m}=0 \quad \frac{y_1+y_2+y_3}{3}=10\) \(\frac{20}{ m }=30\) \(m =2 / 3\) \(\text { and } c-m=6\) \(c=\frac{2}{3}+6 \Rightarrow \frac{20}{3}=c\)…
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