What can we say if a normal to an ellipse passes through this point?

What can we say if a normal to an ellipse passes through this point?

Question

Prove that the equation of the normal to the ellipse $x^2/a^2 + y^2/b^2=1$ at the point $P\,(a\cos\theta,b\sin\theta)$ is $\frac{ax}{\cos \theta} - \frac{by}{\sin \theta} = a^2-b^2.$

If there is a value of $\theta$ between $0$ and $\pi/2$ such that the normal at $P$ passes through one end of the minor axis, show that the eccentricity of the ellipse must be greater than $1/\sqrt{2}$.

Suggestion

Prove that the equation of the normal to the ellipse $x^2/a^2+y^2/b^2=1$ at the point $P\,(a\cos\theta,b\sin\theta)$ is $\frac{ax}{\cos \theta} - \frac{by}{\sin \theta} = a^2-b^2.$

If $x = f(t), y = g(t)$, what is $\dfrac{dy}{dx}$ in terms of $t$?

Or could we use implicit differentiation?

You might use this applet to explore the problem.

[interactive embed (geogebra-iframe): //www.geogebratube.org/material/iframe/id/ew26SakK/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.

Download GeoGebra file

What is the eccentricity of an ellipse?

The eccentricity, $e$, is a measure of how “squashed” the ellipse is. It can be defined in terms of the major and minor axes. $e = \sqrt{1-\dfrac{b^2}{a^2}}$

Solution

Prove that the equation of the normal to the ellipse $x^2/a^2+y^2/b^2=1$ at the point $P\,(a\cos\theta,b\sin\theta)$ is $\frac{ax}{\cos \theta} - \frac{by}{\sin \theta} = a^2-b^2.$

Implicitly differentiating

the equation of the ellipse with respect to

$x$

, we find

$$\frac{2x}{a^2} + \frac{2y}{b^2} \frac{dy}{dx} = 0$$

and therefore

$$\begin{align*} \frac{dy}{dx} &= - \frac{b^2x}{a^2y} \\ &= - \frac{b^2 a \cos \theta}{a^2 b \sin \theta} \\ &= - \frac{b \cos \theta}{a \sin \theta}. \end{align*}$$

Alternatively, we could say $x=a\cos\theta, y=b\sin\theta$ which gives on differentiating $\dfrac{dx}{d\theta}=-a\sin \theta, \dfrac{dy}{d\theta}=b\cos \theta$.

Now dividing these gives us $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}=-\frac{b\cos\theta}{a\sin\theta}.$

Since the normal is perpendicular to the tangent, it has slope $m = -\frac{1}{dy/dx} = \frac{a \sin \theta}{b \cos \theta}.$

Hence the equation of the line is $y - b\sin\theta = \frac{a \sin \theta}{b \cos \theta}\left(x - a \cos \theta\right)$ which we can rearrange to obtain $a^2 - b^2 = \frac{ax}{\cos \theta} - \frac{by}{\sin\theta}$ as required.

The minor axis is the shortest diameter of the ellipse (a diameter of the ellipse is a chord that passes through the centre).

The ends have coordinates $(0,\pm b)$ (assuming that $a \geq b>0$, which we can say without loss of generality).

If we know there is a value of $\theta$ between $0$ and $\pi/2$ such that the normal at $P$ passes through one end of the minor axis, then from the diagram, this normal must pass through $(0, -b)$.

Substituting $(0,-b)$ into the equation of the normal gives $a^2 - b^2 = \frac{b^2}{\sin\theta} \qquad \implies \qquad \left(\frac{a}{b}\right)^2 = 1 +\frac{1}{\sin\theta}.$

But now since $0 < \sin\theta < 1$ we conclude $\left(\frac{a}{b}\right)^2 > 2 \qquad \implies \qquad \frac{b^2}{a^2} < \frac{1}{2}.$

If an ellipse is a ‘squashed’ circle, then the eccentricity measures the degree of ‘squashedness’. We have the standard result that $e = \sqrt{1-\dfrac{b^2}{a^2}}$.

In our case, the eccentricity $e = \sqrt{1-\frac{b^2}{a^2}} > \sqrt{1 - \frac 1 2} = \frac{1}{\sqrt{2}}$ as required.