Question

At any point $P$ on the parabola $y^2-2y-4x+5=0$, a tangent is drawn which meets the directrix at $Q.$ The locus of the points $P$ which divides $QP$ externally in the ratio $\frac{1}{2}:1$ , is

• A. $(x+1)(1-y^2)+4=0$
• B. $x+1=0$
• C. $(1-y^2)-4=0$
• D. None of these

Answer 1

Answer: (A)

$(x+1)(1-y^2)+4=0$

${(y-1)}^2=4(x-1).$

$P$ has coordinates $x=1+t^2,y=1+2t.$

Tangent at $P$ is $(x-1)-(y-1)t+t^2=0$

Directrix is $x=0$

$\therefore Q=[0,t+1-\frac{1}{t}]$

If $R(x,y)$ divides $QP$ externally in the ratio $1:2,$ then

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

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

or $(x+1){(1-y)}^2+4=0.$