# Number of normals from a point to an ellipse

From a typical point $P$ inside an ellipse, how many points $Q_i$ on the ellipse have $PQ_i$ normal to the ellipse? Someone asked me at school many years ago but I don't think I worked it out.

For every ellipse $$\mathcal{E}$$, there is a curve called ellipse evolute $$^\color{blue}{[1]}$$ associated with it. The ellipse evolute is the locus of centers of curvature $$^\color{blue}{[2]}$$ for $$\mathcal{E}$$. It is also the envelope of the normal lines of $$\mathcal{E}^\color{blue}{[2]}$$.

For a point $$P$$ inside $$\mathcal{E}$$, the number of points $$Q$$ on $$\mathcal{E}$$ where $$PQ$$ normal to $$\mathcal{E}$$ can be either $$2, 3$$ or $$4$$. It depends on whether $$P$$ is lying outside, "on" or inside the ellipse evolute.

There are exceptions for the "on" case. When $$P$$ is one of the $$4$$ cusps, the number of normals is $$2$$ instead of $$3$$.

The graph below illustrates what happens when $$P$$ lies on the ellipse evolute (the black star shaped curve) and the three $$Q$$ such that $$PQ$$ normal to $$\mathcal{E}$$.

Following is a brief analysis of the problem. For an alternate and more complete treatment, take a look at the paper Apollonius' ellipse and evolute revisited by Frederick Hartmann and Robert Jantzen.

Let use choose a coordinate system such that $$\mathcal{E}$$ is given by the equation

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$ where $$a > b > 0$$ are the semi-major and semi-minor axis of $$\mathcal{E}$$. Let $$c = \sqrt{a^2-b^2}$$, the two foci of $$\mathcal{E}$$ are located at $$(\pm c,0)$$.

Let $$P = (p,q)$$ be a point inside $$\mathcal{E}$$ and $$Q = (x,y)$$ be a point on $$\mathcal{E}$$. It is known that the normal of $$\mathcal{E}$$ at $$Q$$ is along the direction $$(\frac{x}{a^2},\frac{y}{b^2})$$. The condition for $$PQ$$ normal to $$\mathcal{E}$$ is given by

$$\frac{x}{a^2} : \frac{y}{b^2} = x - p : y - q \quad\iff\quad \frac{x}{a^2}(y-q) - \frac{y}{b^2}(x-p) = 0$$

This is the equation for a hyperbola $$\mathcal{H}$$ ( the orange hyperbola in above graph ) with $$P$$ lying on it. Since $$P$$ is inside $$\mathcal{E}$$ and the two arms of the branch of $$\mathcal{H}$$ holding $$P$$ extends to infinity. Each of the arm will intersect $$\mathcal{E}$$ at least once. This means $$\mathcal{E}$$ and $$\mathcal{H}$$ intersected at least twice. Since five points determine a conic, the number of intersections between $$\mathcal{E}$$ and $$\mathcal{H}$$ is at most $$4$$.

To determine the actual number of intersections, let us introduce a new coordinate system

$$(u,v) = (\frac{x}{a},\frac{y}{b})\quad\iff\quad (x,y) = (au, bv)$$ In terms of $$(u,v)$$, the ellipse becomes the unit circle $$u^2 + v^2 = 1$$ and the hyperbola $$\mathcal{H}$$ becomes

$$uv + \tilde{q} u - \tilde{p} v = 0\quad\text{ where }\quad \begin{cases}\tilde{p} = \frac{ap}{c^2}\\ \tilde{q} = \frac{bq}{c^2}\end{cases}$$ Consider the rational parametrization of the circle $$t\quad \mapsto\quad (\frac{1-t^2}{1+t^2}, \frac{2t}{1+t^2})$$ The condition that a point on the circle intersect $$\mathcal{H}$$ becomes

\begin{align} & \left(\frac{1-t^2}{1+t^2}\right)\left(\frac{2t}{1+t^2}\right) + \tilde{q}\left(\frac{1-t^2}{1+t^2}\right) - \tilde{p}\left(\frac{1-t^2}{1+t^2}\right) = 0\\ \\ \iff & \tilde{q}t^4 + 2 (\tilde{p}+1) t^3 + 2(\tilde{p}-1) t - \tilde{q} = 0 \end{align} When $$P$$ doesn't lie on the $$x$$-axis, $$\tilde{q} \ne 0$$ and last expression is a quartic equation. It "usually" have either two or four real roots. When we changes the position of $$P$$, the number of roots jumps at those place where the discriminant of the quartic polynomial vanishes.

By brute force computation, the discriminant of the quartic polynomial is given by $$\Delta(\tilde{p},\tilde{q}) = -256\left[ (\tilde{p}^2+\tilde{q}^2-1)^3 + 27\tilde{p}^2\tilde{q}^2\right]$$

We have 3 possible cases:

1. When $$P$$ is near $$O$$, $$\Delta(\tilde{p},\tilde{q}) > 0$$. The quartic equation has either $$4$$ or $$0$$ real roots. Since we know $$\mathcal{H}$$ intersect $$\mathcal{E}$$ at least twice, there are four $$Q$$ on $$\mathcal{E}$$ such that $$PQ$$ is normal to $$\mathcal{E}$$.

2. When $$P$$ is far away from $$O$$, $$\Delta(\tilde{p},\tilde{q}) < 0$$. The quartic equation has only $$2$$ real roots and hence there are only two $$Q$$ that make $$PQ$$ normal to $$\mathcal{E}$$.

3. On the special case $$\Delta(\tilde{p},\tilde{q}) = 0$$, the quartic equation "usually" have 3 distinct real roots. One of them is a double root which corresponds to $$\mathcal{H}$$ is tangent to $$\mathcal{E}$$ at some points. There are three $$Q$$ that make $$PQ$$ normal to $$\mathcal{E}$$.

There are exceptions. When $$P$$ lies on the $$y$$-axis, $$\tilde{p} = 0$$ and $$\Delta(\tilde{p},\tilde{q}) = 0 \implies \tilde{q} = \pm 1$$. The quartic equation now have a triple root and the number of real roots is $$2$$ instead of $$3$$. This corresponds to $$P$$ is one of the $$4$$ cusps of the evolute.

As mentioned before, the picture above illustrates the $$3^{rd}$$ case. The black star shaped curve is the ellipse evolute where $$\Delta(\frac{ap}{c^2},\frac{bq}{c^2}) = 0$$. When $$P$$ is lying on it, one branch of the hyperbola $$\mathcal{H}$$ will be touching the ellipse $$\mathcal{E}$$. If you move $$P$$ inside the ellipse evolute, the bottom branch of $$\mathcal{H}$$ will move inwards too and $$\mathcal{H}$$ will start to intersect $$\mathcal{E}$$ at four places.

To obtain a simpler expression for the ellipse evolute, let $$r = |\tilde{p}|^{\frac23}, s = |\tilde{q}|^{\frac23}$$, we have \begin{align} \Delta(\tilde{p},\tilde{q}) = 0 \iff & (r^3 + s^3 - 1)^3 + 27r^3s^3 = 0\\ \iff & r^3 + s^3 - 1 + 3rs = 0\\ \iff & (r+s-1)(s^2-rs+s+r^2+r+1) = 0\\ \iff & (r+s-1)\left((s+r+2)^2+3(s-r)^2\right) = 0\\ \iff & r + s = 1\\ \iff & |ap|^{\frac23} + |bq|^{\frac23} = c^{\frac43} \end{align} Reproducing the equation for ellipse evolute commonly appear in the literature.

Notes

$$\color{blue}{[1]}$$ ellipse evolute is a special case of a kind of curve called astroid.

$$\color{blue}{[2]}$$ The wiki page of evolute has the definition of center of curvature. It also has a nice animation showing the ellipse evolute as an envelop of the normals.

• Thankyou for the explanation. I remember concave diamonds, but I don't think I got this far before. Dec 19, 2013 at 14:57