Question

A variable chords of the parabola $y^2=8x$ touches the parabola $y^2=2x$. The locus of the point of intersection of the tangent at the end of the chord is a parabola. Find its latus rectum.

Answer 1

$y^2=8x=4a_{1}x\Rightarrow a_1=2$

$y^2=2x=4a_{2}x\Rightarrow a_2=\frac{1}{2}$

AB is chord of contact for pt 'P' wrt $y^2=8x$

Equation of AB: $T=0$

$\Rightarrow ky=8\left(\frac{x+h}{2}\right)$

$\Rightarrow ky=4x+4h$

$\Rightarrow 4x-ky+4h=0$ …….$\left(1\right)$

AB is tangent to $y^2=2x$

let $Pt(x,y)=(a_2{t}^2,2a_{2}t)$

Tangent at $\left(t\right)\Rightarrow y=\frac{x}{r}+at$

$y=\frac{x}{t}+a_{2}t$

$=\frac{x}{t}+\frac{t}{2}$

$\Rightarrow \frac{x}{t}-y+\frac{t}{2}$ ……$\left(2\right)$

$\left(1\right)$ & $\left(2\right)$ represents same time

$\frac{x}{t}=4\Rightarrow t=y_4$

$-ky=-y\Rightarrow K=1$

$4h=\frac{t}{2}\Rightarrow h=\frac{1}{32}$

Compare $1$ & $2$

$\frac{4}{(Yt}=\frac{-k}{-1}=\frac{4h}{\frac{t}{2}}$

$\Rightarrow 4t-k$ & $k=\frac{8h}{t}$

$kt=8h$

$k.\frac{k}{4}=8h$

$k^2=32h$

$\therefore y^2=32x$

Now $y^2=32x=4Ax$

$A=8$

L.R$=4A=32\Rightarrow 32$.