Question

$PQ$ is a chord of parabola which subtends right angle at vertex. Then locus of centroid of triangle $PSQ$, where $S$ is the focus of given parabola , is

• A. $x^2=4(y+3)$
• B. $x^2=\frac{4}{3}(y-3)$
• C. $x^2=\frac{-4}{3}(y+3)$
• D. $x^2=\frac{4}{3}(y+3)$

Answer 1

Answer: (B)

$x^2=\frac{4}{3}(y-3)$

Given $PO$ is perpendicular to $OQ$

$SlopeofOP=\frac{t_1^2}{2t_1}=\frac{t_1}{2}$

Similarly, $SlopeofOQ=\frac{t_2^2}{2t_2}=\frac{t_2}{2}$

$SlopeofOP\times SlopeofOQ=-1$

$\frac{t_1}{2}\times \frac{t_2}{2}=-1$

$t_1{t}_2=-4$ ---(i)

Let centroid of $△PSQ$ is $C(h,k)$

Therefore, Centroid, $C(h,k)=(\frac{2t_1+2t_2+0}{3},\frac{t_1^2+t_2^2+1}{3})$

$⟹h=\frac{2t_1+2t_2+0}{3}$ and $k=\frac{t_1^2+t_2^2+1}{3}$

$\frac{3h}{2}=t_1+t_2$ ---(ii) and $t_1^2+t_2^2=3k-1$ ---(iii)

Squaring (ii),

$t_1^2+t_2^2+2t_1{t}_2=\frac{9h^2}{4}$

$3k-1+2(-4)=\frac{9h^2}{4}$

$12k-4-32=9h^2$

$9h^2=12k-36=12(k-3)$

$h^2=\frac{12}{9}(k-3)=\frac{4}{3}(k-3)$

$\therefore$ Locus of Centroid,C is,

$x^2=\frac{4}{3}(y-3)$ -----------Option B