Answer to Question #223992 in Analytic Geometry for Bless

Question

 Show that, if the chord joining the points P(ap² , 2ap), Q(aq² , 2aq) on the parabola y²= 4ax passes through (a, 0), then pq = −1. Further, the tangent at P meets the line through Q parallel to the axis of the parabola at R. Prove that the line x + a = 0 bisects PR.

Answer 1

 The chord joining the points  $P(ap^2, 2ap)$ and $Q(aq^2, 2aq)$

$$y=mx+b$$

$$slope=m=\dfrac{2aq-2ap}{aq^2-ap^2}=\dfrac{2}{q+p}, a\not=0,q\not=\pm p,$$

$$y=\dfrac{2}{q+p}x+b$$

Point  $P(ap^2, 2ap)$

$$2ap=\dfrac{2}{q+p}(ap^2)+b$$

$$b=\dfrac{2(apq+ap^2-ap^2)}{q+p}$$

$$b=\dfrac{2apq}{q+p}$$

$$y=\dfrac{2}{q+p}x+\dfrac{2apq}{q+p}$$

Point $(a, 0)$

$$0=\dfrac{2}{q+p}a+\dfrac{2apq}{q+p}$$

$$pq=-1$$

Therefore if the chord joining the points P(ap² , 2ap), Q(aq² , 2aq) on the parabola y²= 4ax passes through (a, 0), then pq = −1.

The tangent at $P(ap^2, 2ap)$

$$y^2= 4ax$$

$$2yy'=4a$$

$$y'=\dfrac{2a}{y}$$

$$slope=\dfrac{2a}{2ap}=\dfrac{1}{p}$$

$$y-2ap=\dfrac{1}{p}(x-ap^2)$$

$$y=\dfrac{1}{p}x+ap$$

The tangent at P meets the line through Q parallel to the axis of the parabola at R

$$y=2aq$$

$$\dfrac{1}{p}x+ap=2aq$$

$$x=2apq-ap^2$$

$$R(2apq-ap^2, 2aq)$$

$$PR:\dfrac{x_1+x_2}{2}=\dfrac{ap^2+2apq-ap^2}{2}=apq$$

$$\dfrac{y_1+y_2}{2}=\dfrac{2ap+2aq}{2}=ap+aq$$

The middle of $PR$ is the point $(apq, ap+aq).$

If $pq = −1,$ then

$$\dfrac{x_1+x_2}{2}=apq=-a$$

The middle of $PR$ is the point $(-a, ap+aq).$

Then the line $x+a=0$ bisects PR.