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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?

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  • $\begingroup$ You might know that a ray of light emanating from the focus will be reflected in the direction of the parabola axis. So, it certainly won't be reflected back and pass through the focus again, unless it hits the vertex. Therefore, it can not be a normal to the parabola. $\endgroup$ – bubba May 20 '14 at 11:46
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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 $x_0=0$.

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  • $\begingroup$ 50 years ago, I learned in high school that the tangent line at the point $(x_0,y_0)$ on the parabola $x^2 = 4ay$ has the equation $x_0x = 2a(y+y_0)$. This works for any conic section -- you replace "half" of the $x$'s by $x_0$ and half of the $y$'s by $y_0$. More generally, this "half-way" substitution gives you the polar line of the point $(x_0,y_0)$, regardless of whether it lies on the conic. Do they still teach this stuff? $\endgroup$ – bubba May 20 '14 at 11:41

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