Question

 Prove that the line x − 2y + 4a = 0 touches the parabola y²= 4ax, and find the coordinates of P, the point of contact. If the line x − 2y + 2a = 0 meets the parabola in Q, R, and M is the mid-point of QR, prove that PM is parallel to the axis of x, and that this axis and the line through M perpendicular to it meet on the normal at P to the parabola.

Answer 1

$$x − 2y + 4a = 0$$

$$y^2= 4ax$$

Then

$$y^2=4a(2y-4a)$$

$$y^2-8ay+16a^2=0$$

$$(y-4a)^2=0$$

$$y=4a$$

Since the equation $y^2=4a(2y-4a)$ has the only root, then the line $x − 2y + 4a = 0$ touches the parabola $y^2=4ax.$

$$y=4a, x=(4a)^2/4a=4a$$

The point of contact $P(4a, 4a).$

Suppose that $Q$ has coordinates $(aq^2, 2aq)$ and $R$ has coordinates $(ar^2, 2ar).$

The line $x − 2y + 2a = 0$ meets the parabola in $Q, R$

$$aq^2-4aq+2a=0$$

$$q^2-4q+2=0, a\not=0$$

$$q=2\pm\sqrt{2}$$

$$ar^2-4ar+2a=0$$

$$r^2-4r+2=0, a\not=0$$

$$r=2\pm\sqrt{2}$$

$$(2-\sqrt{2})^2=6-4\sqrt{2}$$

$$(2+\sqrt{2})^2=6+4\sqrt{2}$$

Suppose that $Q$ has coordinates $(a(6-4\sqrt{2}), a(4-2\sqrt{2}))$ and $R$ has coordinates $(a(6+4\sqrt{2}), a(4+2\sqrt{2})).$

Since $M$ is the mid-point of $QR,$ $M$ has coordinates $( 6a, 4a).$

Then the equation of $PM$ is $y=4a.$

Hence $PM$ is parallel to the axis of $x.$

Find the equation of the normal at $P$ to the parabola

$$slope=\dfrac{-1}{\dfrac{1}{2}}=-2$$

$$y=-2x+b$$

$$4a=-2(4a)+b=>b=12a$$

$$y=-2x+12a$$

The lines $y=0, x=6a,$ and $y=-2x+12a$ meet at the point $(6a, 0).$