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# locus of the point of intersection of tangent to $y^2=4ax$ which intercept a constant length d on the directrix

Show that the locus of the point of intersection of tangent to $y^2=4ax$ which intercept a constant length d on the directrix is $(y^2+4ax) (x+a^2)^2= d^2x^2$

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A)
Let P be the point of intersection of the tangents to the parabola.
Hence the equation of the pair of tangents $PQ$ and $PR$ is
$\bigg[(y y_1-2a(x+x_1)\bigg]^2 =(y^2-4ax)(y_1^2-4ax_1)$
These lines meet the directix $x+1=0$ or $x=-a$.
Substituting this we get,
$\bigg[(yy_1-2a(-a+x_1)\bigg]^2 =(y^2+4a^2)(y_1^2- 4ax_1)$
On expanding we get ,
$(yy_1)^2+[(2a(x_1-a)]^2-4ayy_1(x_1-a)=(yy_1)^2-4ax_1y^2+4a^2y_1^2 -16 a^3x_1$
(ie) $y^2 [y_1^2-y_1^2+4ax_1] -y[4a(x_1-a)y_1]+[4a^2(x_1-a)^2-4a^2(y_1^2-4ax_1)]y=0$
$x_1y^2-y_1(x_1-a)y+a[(x_1-a)^2-(y_1^2-4ax_1)]=0$
=> $x_1 y^2-yy_1 (x_1-a)+a[(x_1+a)^2-y_1^2]=0$
Here $y_1$ and $y_2$ are the ordinates of the point of intersection of tangents with directrix $x+a=0$
Sum of the ordinates is $y_1+y_2 = \large\frac{-b}{a}$
$\qquad= \large\frac{(x_1-a)y_1}{x_1}$
Product of the ordinater is
$y_1y_2= \large\frac{a[(x_1+a)^2-y_1^2]}{x_1}$
$\therefore d^2 =(y_1-y_2)^2=(y_1+y_2)^2 -4y_1y_2$
Now substituting the values we get,
$d^2= \large\frac{(x_1 -a)^2y_1^2 - 4ax_1[(x_1+a)^2-y_1^2]}{x_1^2}$
$x_1^2d^2 = (x_1-a)^2 y_1^2 -4ax_1 [(x_1+a)^2-y_1^2]$
$x_1^2d^2 = y_1^2\bigg[(x_1-a)^2+4ax_1\bigg] -4ax_1 (x_1+a)^2$
$\qquad= y_1^2(x_1+a)^2-4ax_1(x_1+a)^2$
$\therefore x_1^2d^2= (x_1+a)^2(y_1-4ax_1)$
Hence the locus of $(x_1,y_1)$ is
$x^2d^2=(y-4ax)(x+a)^2$