Question

Let PQ be a variable focal chord of the parabola $y^2=4ax$ where vertex is A. Locus of , centroid of triangle APQ is a parabola $P_1$

• A. $\frac{2a}{3}$
• B. $\frac{4a}{3}$
• C. $\frac{8a}{3}$
• D. $\frac{16a}{3}$

Answer 1

The correct option is B $\frac{4a}{3}$

Let $P$ will be demanded are

$P\left(t\right)=P(a{t}^2,2at)$

& $Q(t^1,2a{t}^1)$

For focal chord end points

$t{t}^1=-1$

$\therefore t^1=-\frac{1}{t}$

$\therefore Q(-\frac{1}{7})=Q\left(\frac{a}{t^2,-\frac{2a}{t}}\right)$Let $G(\alpha ,\beta )$ be centroid of $△APQ$ :

$\alpha =\frac{a{t}^2+\frac{o}{t^2}}{3}+o$

$\beta =\frac{2at+(-\frac{2o}{t})+o}{3}$

$\frac{3\alpha }{a}=t^2+\frac{1}{t^2}$ $\frac{3\beta }{2a}=(t-\frac{1}{t})$

$\frac{3\alpha }{a}=(t^2+\frac{1}{t^2}-2)+2$

$\frac{3\alpha }{\alpha }={(t-\frac{1}{t})}^2+2$

$\frac{3\alpha }{a}={\left(\frac{3\beta }{2a}\right)}^2+2$

$12\alpha a=9{\beta }^2+8a^2$

Replacing $\alpha$ by $x$ & $\beta$ by $y$

$9y^2=12ax-8a^2$

$y^2=\frac{4ax}{3}-\frac{8a^2}{9}$

$y^2=\frac{4a}{3}[x-\frac{2a}{3}]$

The locus is a parabola with vertex at $(=\frac{2a}{3},o)$ & locus reacts length is $\frac{4a}{3}$