Answer to Question #224001 in Analytic Geometry for Bless

Question

The coordinates of the ends of a focal chord of the parabola y²= 4ax are (x₁, y₁) and (x₂, y₂). Show that x₁x₂ = a² and y₁y₂ = −4a².

Answer 1

Answer:-

Let $P(x_1,y_1) = (at_1^2 , 2at_1)$ and $Q(x_2, y_2) = (at_2^2 , 2at_2)$ be two end points of a focal chord. P, S, Q are collinear.

Slope of PS = Slope of QS

$\dfrac{2at_1}{at_1^2-a}=\dfrac{2at_2}{at_2^2-a}\\ t_1t_2^2-t_1=t_2t_1^2-t_2\\ t_1t_2(t_2-t_1)+(t_2-t_1)=0\\ t_1t_2=-1$

from (1)

$x_1x_2=at_1^2at_2^2=a^2(t_1t_2)^2=a^2\\ y_1y_2=2at_12at_2=4a^2(t_1t_2)=-4a^2\\$