How many normals to a parabola pass through $P$?

How many normals to a parabola pass through $P$?

Question

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

Prove that, if $p^2>8$, two chords can be drawn through the point $(ap^2,2ap)$ which are normal to the parabola at their second points of intersection, and that the line joining these points of intersection meets the axis of the parabola in a fixed point, independent of $p$.

Suggestion

We need to find the gradient of the normal at this point. How could we find the gradient of the curve itself?

Can we find values of $t$ such that $(ap^2,2ap)$ lies on the normal at $T$? Can we get a cubic equation for $t$ in terms of $p$? Do we know any of its roots?

You might find the following applet helpful in visualising the normals to the curve.

[interactive embed (geogebra-iframe): //www.geogebratube.org/material/iframe/id/xryG3d7n/width/700/height/400/border/888888/rc/false/ai/false/sdz/true/smb/false/stb/false/stbh/false/ld/false/sri/true/asb/false/at/html5] — not part of the text import; see the source page.

Download GeoGebra file

As we vary $p$ using the slider, the point $P = (ap^2, 2ap)$ moves around the curve. All possible normals through $P$ are shown, in green and purple.

Solution

Approach 1: differentiate with respect to the parameter $t$.

Differentiating with respect to $t$, we see $\dfrac{dy}{dx} = \dfrac{dy/dt}{dx/dt}= \dfrac{2a}{2at}= \dfrac{1}{t}$.

So the gradient of the normal at $T$ is $-t$.

The point $(at^2,2at)$ is on the line which gives us its equation as $y = -tx +at^3+2at$.

Approach 2: find $y$ directly

We have $y=2\sqrt{ax}$ (taking the positive square root for the top half of the curve).

Now we can differentiate to find $\dfrac{dy}{dx} = 2\cdot\dfrac{1}{2}(ax)^{-1/2}a= \sqrt{\dfrac{a}{x}}$, and at $x=at^2$ (noting that $t$ is positive), this gives us $\dfrac{dy}{dx} = \dfrac{1}{t}$. As above, we conclude the normal has gradient $-t$.

We get the same result for the bottom half of the curve.

Approach 3: implicit differentiation

Differentiating $y^2=4ax$ with respect to $x$ gives, using the chain rule, $2y \dfrac{dy}{dx} = 4a$ and therefore the gradient at $(at^2,2at)$ is $\dfrac{4a}{2y} = \dfrac{4a}{4at}=\dfrac{1}{t}$.

We seek points $T = (at^2, 2at)$ where the normal at $T$ passes through $P = (ap^2, 2ap)$.

So we need $(ap^2,2ap)$ to lie on the line $y = at^3+2at-xt$.

Substituting in these values of $(x,y)$ we obtain $2ap = at^3 + 2at-ap^2t$ or (cancelling the $a$) $t^3 + (2-p^2)t - 2p = 0.$

We already know one solution: $p=t$ (because the normal at the point given by $p$ passes through the curve at this same point.)

So we can take out from this cubic a factor of $(p-t)$ to give $t^2 + pt + 2 = 0.$

A quadratic has two roots if and only if its discriminant is positive, or “$b^2-4ac>0$”, which here gives simply $p^2-8>0$, as required.

Finally, we consider the line joining $T_1=(at_1^2,2at_1)$ and $T_2=(at_2^2,2at_2)$ on the parabola (where $t_1$ and $t_2$ denote the two roots of the quadratic).

We want to find where this intersects the axis of the parabola, the line $y=0$.

The gradient of the line is given by ‘the difference in $y$ coordinates divided by the difference in $x$ coordinates’, which is $\frac{2a(t_2-t_1)}{a(t_2^2-t_1^2)} = \frac{2}{t_1+t_2}.$

Hence the equation of the line is $y - 2at_1 = \frac{2}{t_1+t_2} (x - a t_1^2),$ and by setting $y=0$ we find $x = at_1^2 + \frac{1}{2} (t_1+t_2)(-2at_1) = -at_1t_2.$

We can recall now that the product of the roots of the quadratic $x^2 + ax + b = 0$ is given by $b$, the constant term.

But in $t^2+pt+2=0$ the constant term is simply $2$, giving the $x$-intercept to be $x = -2a$, which is indeed independent of $p$.