Question

Find the locus of the points of intersection of tangents drawn at the ends of all normal chords of the parabola $y^2=8\left(x-1\right)$

Answer 1

The given parabola equation is $Y^2=8X$ where $Y=y$ and $X=x-1$

Every point on this parabola is $(x=2t^2,y=4t)=(2t^2+1,4t)$

Normal at $(2t^2+1,4t)$ is

$tX+Y=2\left(2t\right)+2t^3$

$\Rightarrow t(x-1)+y=4t+2t^3$ .........$\left(1\right)$

Suppose the tangents at the ends of normal chord intersect at $P(x_1,y_1)$. Then, the normal chord is the chord of contact of $P(x_1,y_1)$ and hence its equation is

$Y{y}_1-4(X+x_1)=0$

$\Rightarrow Y{y}_1-4(x-1+x_1)=0$

$\Rightarrow Y{y}_1-4x+4-4x_1=0$ .......$\left(2\right)$

Equations $\left(1\right)$ and $\left(2\right)$ represent the same straight line.

$\therefore \frac{t}{-4}=\frac{1}{y_1}=\frac{4t+2t^3}{4x_1-4}$

$\Rightarrow t=-\frac{4}{y_1}and-1=\frac{4+2t^2}{x_1-1}$

$\Rightarrow -(x_1-1)=4+2\left(\frac{16}{{y_1}^2}\right)$

$\Rightarrow -x_1-3=\frac{32}{{y_1}^2}$

$\Rightarrow {y_1}^2(x_1+3)+32=0$

Hence the locus at $(x_1,y_1)$ is

$y^2(x+3)+32=0$