Question

Let $PQ$ be the chord of the parabola $y^2=4x$ with slope $=2$. Locus of the point $R$ which divides the chord $PQ$ intemally in the ratio $1:2$ is a parabola whose length of latus rectum is $L$. Find the value of $18L$.

Answer 1

Answer:

$\text{ Given }\frac{2}{t_1+t_2}=2$
$\therefore t_1+t_2=1$ ....(1)
$\text{ Also }3h=t_1^2+2t_2^2$ ....(2)
$3k=2t_1+4t_2$
$3k=2\left(t_1+2t_2\right)$ .....(3)
$\text{ Now, }3k=2\left(1+t_2\right)$
$\frac{3k}{2}-1=t_2$
$\therefore t_1=1-t_2=2-\frac{3k}{2}$
$3h=t_1^2+2t_2^2$
$3h={\left(2-\frac{3k}{2}\right)}^2+2{\left(\frac{3k}{2}-1\right)}^2$
$3h=4+\frac{9k^2}{4}-6k+\frac{9k^2}{2}+2-6k$
$3h=\frac{27k^2}{4}-12k+6$
$3(x-2)=\frac{27}{4}\left(y^2-\frac{16y}{9}\right)$
$(x-2)=\frac{9}{4}\left[{\left(y-\frac{8}{9}\right)}^2-\frac{64}{81}\right]$
$\frac{4}{9}(x-2)={\left(y-\frac{8}{9}\right)}^2-\frac{64}{81}$
$\therefore {\left(y-\frac{8}{9}\right)}^2=\frac{4}{9}\left(x-\frac{2}{9}\right)$
hence length of latus recturm $=\frac{4}{9}=L$
$\Rightarrow 18L=8$.