Review question

# When do these normals to a parabola meet on the parabola? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9809

## Solution

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

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 $t_1t_2=2, ...$

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.

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.

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 $O$.

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.