Question

Prove that the locus of the points of intersection of tangents to $y^2=4ax$ which intercept a constant length $d$ on the directrix is $(y^2-4ax)(x+a{)}^2=d^2{x}^2$

Answer 1

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 ${\left[\right(y{y}_1-2a(x+x_1)\left)\right]}^2=(y^2-4ax)(y_1^2-4a{x}_1)$

These lines meet the directrix $x+1=0$ or $x=-a$

Substituting this we get

${\left[\right(y{y}_1-2a(-a+x_1)\left)\right]}^2=(y^2+4a^2)(y_1^2-4a{x}_1)$

On expanding we get

$(y{y}_1{)}^2+[2a(x_1-a{)}^2]-4ay{y}_1(x_1-a)=(y{y}_1{)}^2-4a{x}_1{y}^2+4a^2{y}_1^2-16a^3{x}_1$

$\Rightarrow y^2(y_1^2-y_1^2+4a{x}_1)-y\left[4a\right(x_1-a\left)y_1\right]+\left[4a^2\right(x_1-a{)}^2-4a^2(y_1^2-4a{x}_1)]y=0$

$\Rightarrow x_1{y}^2-y_1(x_1-a)y+a\left[\right(x_1-a{)}^2-(y_1^2-4a{x}_1)]=0$

$\Rightarrow x_1{y}^2-y_1(x_1-a)y+a\left[\right(x_1+a{)}^2-y_1^2]=0$

Here $y_1,y_2$ are ordinates of the point of intersection of tangents with directrix $x+a=0$

Sum of the ordinates is $y_1+y_2=\frac{-b}{a}=\frac{y_1(x_1-a)}{x_1}$

Product of the ordinates is $y_1{y}_2=\frac{c}{a}=\frac{a\left[\right(x_1+a{)}^2-y_1^2]}{x_1}$

$\therefore d^2=(y_1-y_2{)}^2=(y_1+y_2{)}^2-4y_1{y}_2$

Now substituting the values we get

$d^2=\frac{(x_1-a{)}^2{y}_1^2-4a{x}_1[(x_1+a{)}^2-y_1^2]}{x_1^2}$

$\Rightarrow x_1^2{d}^2=(x_1-a{)}^2{y}_1^2-4a{x}_1[(x_1+a{)}^2-y_1^2]$

$\Rightarrow x_1^2{d}^2=y_1^2\left[\right(x_1-a{)}^2+4a{x}_1]-4a{x}_1(x_1+a{)}^2$

$\Rightarrow x_1^2{d}^2=y_1^2(x_1+a{)}^2-4a{x}_1(x_1+a{)}^2$

$\therefore x_1^2{d}^2=(y_1^2-4a{x}_1)(x_1+a{)}^2$

Hence the locus of $(x_1,y_1)$ is $x^2{d}^2=(y^2-4ax)(x+a{)}^2$