Answer to Question #226667 in Analytic Geometry for Anuj

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

Say point A (x₁, y₁) = $(ap_1^2, 2ap_1)$ , and

point B (x₂, y₃) = $(ap_2^2, 2ap_2)$ coordinates of the ends of a focal chord of the

parabola $y^2 = 4ax$

Then O, A and B are collinear, where O is the vertex.

Therefore, Slope of OA = Slope of OB

Or, $\frac{2ap_1}{(ap_1^2-a)} = \frac{2ap_2}{(ap_2^2-a)}$

On solving,, we get

$p_1p_2^2 - p_1 = p_1^2p_2 - p_2$

$p_1 p_2(p_2-p_1) = -(p_2-p_1)$

Or

$p_1 p_2 = -1$

Now

$x_1 x_2 = ap_1^2 . ap_2^2$

Or

$x_1 x_2 = a^2 (1) = a^2$

Therefore,

$x_1 x_2 = a^2 (Proved)$

Now

$y_1 y_2 = 2ap_1 2a p_2 = 4a^2p_1p_2 = -4a^2$

Therefore,

$y_1 y_2 = -4a^2 (Proved)$