Question

A tangent is drawn to parabola $y^2-4x+4=0$ at a point $P$ which cuts the directrix at the point $Q$. If a point $R$ is such that it divides $QP$ externally in ratio $1:2$, then the locus of point $R$ is

• A. $y(x+1{)}^2+4=0$
• B. $y^2(x+1)+4=0$
• C. $y=0$
• D. $(x+1)(y-1{)}^2-4=0$

Answer 1

The correct option is B $y^2(x+1)+4=0$


$y^2-4x+4=0\Rightarrow y^2=4(x-1)$
Therefore, we get
vertex : $(1,0)$, Focus : $(2,0)$ and Directrix : $x=0$
Let $P\text{be}(1+t^2,2t)$
$\therefore$ Slope of the tangent $=\frac{1}{t}$
Equation of tangent at $P$ is
$y-2t=\frac{1}{t}(x-(1+t^2\left)\right)\Rightarrow y-2t=\frac{x}{t}-\frac{1+t^2}{t}$

$Q$ lies on the directrix $x=0$
$\therefore$ Coordinates of $Q$ is $(0,t-\frac{1}{t})\text{and}P(1+t^2,2t)$
$\because R$ divides $PQ$ in $1:2$ externally
$\therefore R(x,y)=(\frac{1+t^2}{-1},\frac{2t-2(t-\frac{1}{t})}{-1})$

$\Rightarrow x=-1-t^2$
and $y=\frac{-2}{t}\Rightarrow t=\frac{-2}{y}$
$\Rightarrow x=-1-\frac{4}{y^2}\Rightarrow y^2(x+1)+4=0$