When do these normals to a parabola meet on the parabola?

When do these normals to a parabola meet on the parabola?

Question

This question should be linked to the problem "Finding circles"

Find the equation of the normal to the parabola $y^2=4ax$ at the point $(at^2,2at)$.

The parameters of the points $P$, $Q$ are $t_1$ and $t_2$ respectively. Show that, if $PQ$ passes through the point $(-2a,0)$, then $t_1t_2=2$ and the normals at $P$ and $Q$ to the parabola meet at a point $R$ on the parabola.

If $O$ is the origin, show, by considering the gradients of the sides of the quadrilateral $OPQR$ or otherwise, that the circumcircle of the triangle $PQR$ passes through $O$.

Suggestion

How can we find the gradient of the parabola at a point with parameter $t$?

If we have the gradient of the normal through the point $(at^2,2at)$, what is the equation of the line?

The parameters of the points $P$, $Q$ are $t_1$ and $t_2$ respectively. Show that, if $PQ$ passes through the point $(-2a,0)$, then $t_1t_2=2$…

What is the gradient of the line that passes through $P$ and $Q$? What about its equation?

…and the normals at $P$ and $Q$ to the parabola meet at a point $R$ on the parabola.

Could we write down the equations of the normals at $P$ and $Q$, and find where they intersect?

Aren’t we really being asked to show that $OPQR$ is a cyclic quadrilateral? What do we know about cyclic quadrilaterals?

Solution

Taking the derivative of the equation $y^2=4ax$ with respect to $x$ we find that $2y\frac{dy}{dx}=4a \qquad \Longrightarrow \qquad \frac{dy}{dx}=\frac{2a}{y}=\frac{2a}{2at}=\frac{1}{t}.$ The gradient of normal to the parabola at the point $(at^2,2at)$ is therefore $-t$ since $\frac{1}{t}\times -t=-1.$ The equation of the normal is then $y-2at=-t(x-at^2) \qquad \Longrightarrow \qquad y=-tx+2at+at^3.$

The parameters of the points $P$, $Q$ are $t_1$ and $t_2$ respectively. Show that, if $PQ$ passes through the point $(-2a,0)$, then $t_1t_2=2, ...$

[interactive embed (geogebra-iframe): //www.geogebratube.org/material/iframe/id/JhGyhfHZ/width/600/height/400/border/888888/rc/false/ai/false/sdz/false/smb/false/stb/false/stbh/false/ld/false/sri/true/asb/false/at/html5] — not part of the text import; see the source page.

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The gradient of the line

$PQ$

is

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

The equation of the line through points

$P$

and

$Q$

is then

$$y-2at_1=\frac{2}{t_1+t_2}(x-at_1^2).$$

This will pass through the point

$(-2a,0)$

if and only if

$$\begin{align*} 0-2at_1&=\frac{2}{t_1+t_2}(-2a-at_1^2) \\ &\iff t_1 = \frac{1}{t_1+t_2}(2+t_1^2) \\ &\iff t_1(t_1+t_2) = 2+t_1^2 \\ &\iff t_1 t_2 = 2, \end{align*}$$

as required.

…and the normals at $P$ and $Q$ to the parabola meet at a point $R$ on the parabola.

[interactive embed (geogebra-iframe): //www.geogebratube.org/material/iframe/id/ZQCGfjpf/width/600/height/400/border/888888/rc/false/ai/false/sdz/false/smb/false/stb/false/stbh/false/ld/false/sri/true/asb/false/at/html5] — not part of the text import; see the source page.

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The two normals passing through

$P$

and

$Q$

have the equations

$$\begin{align*} y &=-t_1x+2at_1+at_1^3 \\ y &=-t_2x+2at_2+at_2^3, \end{align*}$$

using the first part of the question. They intersect when

$$\begin{align*} -t_1x+2at_1+at_1^3 &= -t_2x+2at_2+at_2^3 \\ \Longrightarrow \qquad 2a(t_2-t_1)+a(t_2^3-t_1^3) &= x(t_2-t_1) \\ \end{align*}$$

and factorising

$(t_2^3-t_1^3)$

gives

$$\begin{align*} \qquad 2a(t_2-t_1)+a(t_2-t_1)(t_1^2+t_1t_2+t_2^2) &= x(t_2-t_1) \end{align*}$$

Since

$t_1\neq t_2$

, as we are assuming

$P$

and

$Q$

are distinct, we can divide through by

$(t_2-t_1)$

to get

$$x=a(t_1^2+t_1t_2+t_2^2+2).$$

Substituting for

$2=t_1t_2$

we find that

$$\begin{align*} x &=a(t_1^2+2t_1t_2+t_2^2) \\ &= a(t_1+t_2)^2. \end{align*}$$

Using this expression, we can substitute back into one of the original equations for the normal to find

$$\begin{align*} y &= -t_1a(t_1+t_2)^2+2at_1+at_1^3 \\ &=-t_1^3a-2at_1^2t_2-at_1t_2^2+2at_1+at_1^3\\ &=-2a \times 2 \times t_1 - 2 \times at_2+2at_1\\ &= -2a(t_1+t_2). \end{align*}$$

Thus $R=\left(a(t_1+t_2)^2,-2a(t_1+t_2)\right)$, and comparing this to the general point on the parabola $(at^2,2at)$, we can see that $R$ is clearly on the curve.

Since $P = (at_1^2, 2at_1)$, $Q = (at_2^2, 2at_2)$, $R = (a(t_1+t_2)^2, -2a(t_1+t_2))$, and $O = (0, 0)$, then:

$OP$ has gradient $\dfrac{2}{t_1}$, $PQ$ has gradient $\dfrac{2}{t_1+t_2}$, $QR$ has gradient $-\dfrac{2}{t_1}$, while $RO$ has gradient $-\dfrac{2}{t_1+t_2}$.

Thus $OP$ and $QR$ are reflections of each other in a vertical or horizontal line, and $PQ$ and $OR$ are likewise.

The diagram shows that angle $\widehat{PQR}$ is $180^\circ-\alpha -\beta$, and so the angles $\widehat{POR}$ and $\widehat{PQR}$ add to $180^\circ$, and therefore $OPQR$ is a cyclic quadrilateral.

Thus $O$ lies on the circumcircle of $PQR$, as required.