Question

If $P$ and $Q$ are the points of contact of tangents drawn from the point $T$ to $y^2=4ax$ and $PQ$ be a normal of the parabola at $P$, then the locus of the point which bisects $TP$ is

• A. $x+a=0$
• B. $x+2a=0$
• C. $x=0$
• D. $x=1$

Answer 1

The correct option is A $x+a=0$

Given parabola is $y^2=4ax$
Let $P(a{t}_1^2,2a{t}_1)$ and $Q(a{t}_2^2,2a{t}_2)$, then
$T=(a{t}_1{t}_2,a(t_1+t_2\left)\right)$


Let $R(h,k)$ be the mid point of $TP$

$(h,k)=(\frac{a{t}_1{t}_2+a{t}_1^2}{2},\frac{a(t_1+t_2)+2a{t}_1}{2})$
If normal at $P\left(t_1\right)$ intersect the parabola again at $Q\left(t_2\right)$, then
$t_2=-t_1-\frac{2}{t_2}\Rightarrow t_1(t_1+t_2)=-2\Rightarrow t_1^2+t_1{t}_2=-2$

Now,
$h=\frac{-2a}{2}=-a$
Hence, the locus of the midpoint is
$x+a=0$
This is equation of the directrix of the parabola.