Parabola Ques 37
- At any point $P$ on the parabola $y^{2}-2 y-4 x+5=0$ a tangent is drawn which meets the directrix at $Q$. Find the locus of point $R$, which divides $Q P$ externally in the ratio $\frac{1}{2}: 1$.
$(2004,4 M)$
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Solution:
- Given equation can be rewritten as
$(y-1)^{2}=4(x-1)$, whose parametric coordinates are
$$ x-1=t^{2} \quad \text { and } \quad y-1=2 t $$
i.e. $\quad P\left(1+t^{2}, 1+2 t\right)$
$\therefore$ Equation of tangent at $P$ is,
$t(y-1)=x-1+t^{2}$, which meets the directrix $x=0$ at $Q$.
$\Rightarrow \quad y=1+t-\frac{1}{t} \quad$ or $\quad Q \quad 0,1+t-\frac{1}{t}$
Let $R(h, k)$ which divides $Q P$ externally in the ratio
$\frac{1}{2}: 1$ or $Q$ is mid-point of $R P$.
$\Rightarrow \quad 0=\frac{h+t^{2}+1}{2}$ or $t^{2}=-(h+1)$
and $\quad 1+t-\frac{1}{t}=\frac{k+2 t+1}{2}$ or $t=\frac{2}{1-k}$
From Eqs. (i) and (ii), $\frac{4}{(1-k)^{2}}+(h+1)=0$
or $\quad(k-1)^{2}(h+1)+4=0$
$\therefore \quad$ Locus of a point is $(x+1)(y-1)^{2}+4=0$
BETA
