Question

$AP$ is perpandicular to $PB$ . When $A$ is the vetex of the parabola $y^2=4x$, $P$ on the parabola and $B$ is on the axis of the parabola, then the locus of the centriod of $\Delta PAB$ is

• A. $9y^2=2(3x-4)$
• B. $9y^2=2(3x+4)$
• C. $9x^2=2(3y-4)$
• D. $9x^2=2(3y+4)$

Answer 1

The correct option is A $9y^2=2(3x-4)$

We have,

$y^2=4x$

$A$ is vertex. So, the coordinates of $A$ is $(0,0)$.

Let coordinates of point $P$ be $P(t^2,2t)$.

Slope of $AP=\frac{2t-0}{t^2-0}=\frac{2}{t}$

Now, $AP\perp BP$. So,

Slope of $BP=-\frac{t}{2}$

Therefore, equation of $BP$,

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

For point $B$, $y=0$. So,

$-2t=-\frac{t}{2}(x-t^2)$

$x-t^2=4$

$\Rightarrow x=t^2+4$

So, the coordinates of $B$ are $(t^2+4,0)$.

Let coordinates of the centroid be $(h,k)$.

$\Rightarrow h=\frac{0+t^2+t^2+4}{3}$ and $k=\frac{0+2t+0}{3}$

$3h=2t^2+4$ and $t=\frac{3k}{2}$

Substitute $t=\frac{3k}{2}$ in the expression for $h$.

$3h=2{\left(\frac{3k}{2}\right)}^2+4$

$3h=\frac{9k^2}{2}+4$

$6h=9k^2+8$

$9k^2=2(3h-4)$

Substitute $(x,y)$ for $(h,k)$.

$\Rightarrow 9y^2=2(3x-4)$

Hence, this is the required locus.