[Math] Find the points of intersection of two perpendicular tangents to a parabola $y^2=4ax$.

Question

Find the points of intersection of two perpendicular tangents to a parabola $y^2=4ax$.

Any point on the parabola is $(at^2,2at)$. Hence if they intersect at $(x_1,y_1)$
then
$\dfrac{2at-y_1}{at^2-x_1}=\dfrac{1}{t}\implies at^2-ty_1+x_1=0$ since the slope at $(at^2,2at)=\dfrac{1}{t}$.

In order to the tangents to perpendicular $t_1=t_2$.

But how to find the points of intersection between them.Please help

Answer 1

Let $(x,y)=(at^{2},2at)$, then the equation of the tangent at $[t]$ is

$$x-ty+at^{2}=0 \tag{1}$$

If tangents at $[t]$ and $[t']$ are perpendicular, $$tt'=-1$$

Then the equation of another tangent at $[t']$ is $$x+\frac{y}{t}+\frac{a}{t^{2}}=0$$

That is $$t^{2}x+ty+a=0 \tag{2}$$

$(1)+(2)$, $$(1+t^2)x+a(t^2+1)=0$$

$$\fbox{$x=-a$}$$

The locus is the directrix of the parabola which is the limiting case for director circle of an ellipse or a hyperbola.

Useful fact:

Equation of tangent for conics $ax^2+2hxy+by^2+2gx+2fy+c=0$ at the point $(x',y')$ is given by

$$ax'x+h(y'x+x'y)+by'y+g(x+x')+f(y+y')+c=0$$

Also see another answer of mine for your further interest.