Question

$PG$ is the normal at $P$ to the parabola $y^2=4ax$. $G$ is on the axis, $GP$ is produced to $Q$ such that $PQ=GP$, then

• A. Locus of $Q$ is $y^2=16a(x+2a)$
• B. Focus of locus of $Q$ is $(2a,0)$
• C. Latus rectum of locus of $Q$ is $16a$
• D. Locus of $Q$ is $y^2=16a(x-2a)$

Answer 1

Answer: (A), (B), (C)

The correct options are
A Locus of $Q$ is $y^2=16a(x+2a)$
B Focus of locus of $Q$ is $(2a,0)$
C Latus rectum of locus of $Q$ is $16a$

Normal to the given parabola at $P(a{t}^2,2at)$ is
$PG\equiv y+xt=2at+a{t}^3$
for coordinates of $G$, $y=0$
$\therefore G\equiv (2a+a{t}^2,0)$
Let $Q\equiv (h,k)$
$\because P$ is mid point of $GQ$
$\therefore a{t}^2=\frac{2a+a{t}^2+h}{2}\Rightarrow t^2=2+\frac{h}{a}\cdots \left(1\right)2at=\frac{k}{2}\Rightarrow t=\frac{k}{4a}$
Putting the value of $t$ in $\left(1\right)$, we get
$\frac{k^2}{16a^2}=\frac{2a+h}{a}\Rightarrow k^2=16a(h+2a)$
Hence locus of $Q$ will be
$y^2=16a(x+2a)$