I've been having trouble with the following question:
Show that only one normal can pass through the focus of the parabola $x^2=4ay$ and find from which point on the parabola it originates.
There are no solutions to this question, so any hints would be appreciated.
I am a little confused to where it says that there is only one normal which passes through the parabola, since I thought there would be 3 (2 for symmetry plus one at the vertex). Any clarifications for that also?
The parabola given by the equation $x^2=4ay,a\neq0$, can also be represented by
$$y=\frac{1}{4a}x^2.$$
Take the derivative to find the slope of the tangent line at a point; the slope of the normal line will be the negative reciprocal of the slope of the tangent:
$$y^\prime (x)=\frac{1}{2a}x\\ \implies m_{tangent}=y^\prime (x_0)=\frac{x_0}{2a}\\ \implies m_{normal}=-\frac{1}{y^\prime (x_0)}=-\frac{2a}{x_0}$$
The equation for the normal at the point $(x_0,y_0)=(x_0,x_0^2/4a)$ is
$$y-y_0=-\frac{1}{y^\prime (x_0)}(x-x_0),$$
or equivalently,
$$(x_0-x)+y^\prime (x_0)(y_0-y)=0\\ \implies (x_0-x)+\frac{x_0}{2a}\left(\frac{x_0^2}{4a}-y\right)=0.$$
If the normal passes through the focus $(0,a)$, then we must have
$$0=x_0+\frac{x_0}{2a}\left(\frac{x_0^2}{4a}-a\right)\\ =x_0+\frac{x_0}{2a}\cdot\frac{x_0^2-4a^2}{4a}\\ =x_0+\frac{x_0(x_0^2-4a^2)}{8a^2}\\ =\frac{x_0(x_0^2+4a^2)}{8a^2}.$$
Since $0<\frac{x_0^2+4a^2}{8a^2}$ for all $a\neq0$, it follows that we must have $x_0=0$.