Question

If the tangents at two points of a parabola are at right angles, then show that they intersects at a point on the directrix.

Answer 1

Let equation of the parabola is, $y^2=4ax$

Tangent at any point $(x_1,y_1)$ is given by $yy_1 = 2a(x+x_1)$

Let any two points on the parabola be $(at_1^2,2at_1),(at_2^2,2at_2)$

Then equation of the tangent at $(at_1^2,2at_1)$ is given by,

$2at_1y=2a(x+at_1^2)$

$t_1y=(x+at_1^2) \implies y=\frac{1}{t_1}(x+at_1^2)$

Equation of the tangent at point $(at_2^2,2at_2)$ is given by

$2at_2y=2a(x+at_2^2)$

$t_2y=(x+at_2^2) \implies y=\frac{1}{t_2}(x+at_2^2)$

Since, these these tangents are perpendicular then,

$\frac{1}{t_1}\frac{1}{t_2} = -1 \implies \frac{1}{t_2} = -t_1$ (1)

Also these tangents intersect, then point of intersection is

$\frac{1}{t_1}(x+at_1^2)=\frac{1}{t_2}(x+at_2^2)$

from equation (1), I can write,

$(x+at_1^2) = -t_1^2(x+a\frac{1}{t_1^2})$

$(1-t_1^2)x = -a(1-t_1^2)$

$\implies x=-a$

It is equation of the directrix.