Question

Consider a parabola $y^2=4x$, Let $A$ be the vertex of parabola, $P$ be any point on the parabola and $B$ is a point on the axis of parabola, if $PA\perp PB$, then the locus of centroid of $△PAB$ is

• A. $9y^2=-6x-8$
• B. $9y^2=6x+8$
• C. $9y^2=6x-8$
• D. $9y^2=-6x+8$

Answer 1

The correct option is C $9y^2=6x-8$

Here, $A(0,0)$ $P\equiv (a{t}^2,2at)$
Slope of $AP=\frac{2at-0}{a{t}^2-0}=\frac{2}{t}$
$\Rightarrow$ Equation of $BP$ is $y-2t=-\frac{t}{2}(x-t^2)$
for coordinates of $B$, put $y=0$ in above equation
$\therefore 0-2t=-\frac{t}{2}(x-t^2)$
$\Rightarrow x=t^2+4$
$B\equiv (t^2+4,0)$
Centroid $G=(\frac{0+a{t}^2+t^2+4}{3},\frac{0+2at+0}{3})\Rightarrow G=(\frac{0+t^2+t^2+4}{3},\frac{0+2t+0}{3})(\because a=1)\Rightarrow 3x=2t^2+4,3y=2t\Rightarrow 3x=2{\left(\frac{3y}{2}\right)}^2+4\therefore 9y^2=6x-8$