Locating a line perpendicular to an ellipse

Question

At any given point on an ellipse, the line of the perpendicular is easy to calculate; what about the inverse? Given a line of known orientation, at what point on the circumference of an ellipse would it be perpendicular?

Answer 1

Let $$\begin{cases}x=a \cos(t)\\y=b \sin(t)\end{cases}$$ be the classical parametrization of the ellipse.

The tangent vector $\vec{T}$ and the normal vector $\vec{N}$ for the value of parameter $t$ are:

$$\vec{T}\begin{cases}x'(t)= - a \sin(t)\\y'(t)= \ \ \ b \cos(t)\end{cases} \ \implies \ \vec{N}\begin{cases}b \cos(t)\\a \sin(t)\end{cases}$$

Thus it suffices that the slope of $\vec{N}$ : $\dfrac{a \sin(t)}{b \cos(t)}=\dfrac{a}{b} \tan(t)$ is equal to the slope you give: $\dfrac{\sin(\pi/2-x)}{\cos(\pi/2-x)})=\tan(\pi/2-x).$

In other terms: $\tan(t)=\dfrac{b}{a}cotan(x) \ \iff \ t=\tan^{-1}\left(\dfrac{b}{a}cotan(x)\right).$

Answer 2

Given the perpendicular line, you can find the slope of the tangent line by taking the negative reciprocal of the slope of the perpendicular line. Then, once you have the slope of the tangent at a point on the ellipse, find the first derivative of the ellipse and set it equal to that quantity, and then solve. There will be two solutions.

For example, let the axes of the ellipse be $a$ and $b$ and let the line have slope $m$. Then the slope of the tangent at the point is perpendicular to the perpendicular line, and its slope is $-\frac{1}{m}$. Now we must find the first derivative of the equation of an ellipse. The equation of an ellipse takes the form $$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ If we solve for $y$, we get $$y=\pm b\sqrt{1-\frac{x^2}{a^2}}$$ So that $$y'=\pm \frac{bx}{a\sqrt{a^2-x^2}}$$ We set this equal to our tangent slope and solve the equation $$-\frac{1}{m}=\pm \frac{bx}{a\sqrt{a^2-x^2}}$$ Or, because of the "plus or minus", we get $$\frac{1}{m}=\pm \frac{bx}{a\sqrt{a^2-x^2}}$$ Now square both sides to get $$\frac{1}{m^2}=\frac{b^2x^2}{a^2(a^2-x^2)}$$ Multiply both sides by $(a^2-x^2)$ to get $$\frac{1}{m^2}(a^2-x^2)=\frac{b^2}{a^2}x^2$$ $$\frac{a^2}{m^2}-\frac{1}{m^2}x^2=\frac{b^2}{a^2}x^2$$ $$\frac{a^2}{m^2}=(\frac{b^2}{a^2}+\frac{1}{m^2})x^2$$ $$\frac{a^2}{m^2}=(\frac{a^2+b^2m^2}{a^2m^2})x^2$$ $$x=\pm\sqrt{\frac{a^4}{a^2+b^2m^2}}$$ Then once you have these $x$ values, plug it in and solve for the $y$ values, and you should have your points. This is a lot of algebra, so please let me know in the comments if I made a stupid mistake, and I'll fix it.

Answer 3

This problem is trivial for a unit circle: the direction of the normal at the point $(x,y)$ is just the unit vector $\langle x,y\rangle$ and the direction of the corresponding tangent is $\langle-y,x\rangle$. The unit circle can be transformed into an ellipse in standard position by scaling $x$ by a factor of $a$ and $y$ by a factor of $b$. This transformation preserves tangents to the circle/ellipse and parallel lines.

Let the desired direction of the normal be given by the vector $\langle\lambda,\mu\rangle$. (For the angle $x$ from the vertical as shown in your diagram, this would be $\langle\sin x,\cos x\rangle$.) The direction of the corresponding tangent to the ellipse is $\langle-\mu,\lambda\rangle$, which becomes $\langle-\mu/a,\lambda/b\rangle$ in the unit circle’s frame. The corresponding normal is $\langle\lambda/b,\mu/a\rangle$. Normalize and transform back to obtain the points $$\pm\left({a^2\lambda\over\sqrt{a^2\lambda^2+b^2\mu^2}},{b^2\mu\over\sqrt{a^2\lambda^2+b^2\mu^2}}\right).$$ The resemblance to the solution for a circle of radius $r$ is not coincidental.