Just in case the reader is not familiar with the concept of pedal curves, here is a quick definition from Wikipedia:
In mathematics, a pedal curve of a given curve results from the orthogonal projection of a fixed point on the tangent lines of this curve. More precisely, for a plane curve C and a given fixed pedal point P, the pedal curve of C is the locus of points X so that the line PX is perpendicular to a tangent T to the curve passing through the point X.
For clear notation, note that I will use the symbols $\dot x$ and $\dot y$ (Newton's notation) to refer to differentiation regards to $t$ (often interpreted as time), and use the symbols $x'$ and $y'$ if $x$ and $y$ are differentiated with regards to a variable that isn't $t$.
Now, if a curve is represented parametrically by the equations $x = x(t)$ and $y = y(t)$, then the pedal curve with regards to the point $(x_0, y_0)$ can be represented parametrically as:
$$x_p=\frac{\dot x(x_0\dot x + y_0\dot y) + \dot y(x\dot y- y\dot x)}{\dot x^2 + \dot y^2}, \\ y_p=\frac{\dot y(x_0\dot x + y_0\dot y) + \dot x(y\dot x-x\dot y)}{\dot x^2 + \dot y^2}$$
For the ellipse, which has the equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, its pedal curve with regards to the point $(x_0,0)$ can be represented by the parametric equations:
$$x_p = \frac{a^2x_0\sin^2(t)+ab^2\cos(t)}{a^2\sin^2(t)+b^2\cos^2(t)}, \\
y_p = \frac{-abx_0\sin(t)\cos(t)+a^2b\sin(t)}{a^2\sin^2(t)+b^2\cos^2(t)}$$
Now, the area of this pedal curve is:
$$A_p = \int^{2\pi}_0 y_p\dot x_pdt$$
which is where the problem arises. The expression for $\dot x_p$ is already convoluted, so the expression for $y_p\dot x_p$ is even more convoluted. However, the real problem is actually integrating that expression, which is difficult due to the denominator $a^2\sin^2(t)+b^2\cos^2(t)$ and there is no simple way of simplifying the expression.
My textbook tells me that the expression for the area of the pedal curve simplifies down to:
$$\frac{1}{2}\pi(a^2+b^2+x_0^2)$$
so I believe there's something really obvious I'm missing here.
I'm not exactly sure how to proceed with the monstrous integral I have currently on my hands, so I would appreciate any hints.
Best Answer
The tangent at $P = (a \cos(t), b \sin(t))$ is $\frac{\cos(t)}{a}x + \frac{\sin(t)}{b}y = 1$. Hence, the equation of the perpendicular from $X=(x_0, 0)$ to the tangent is $y = \frac{a}{b}\tan(t)(x - x_0)$. If we shift the angle so we're taking it with respect to $X$ instead of the origin, we have $\tan(\theta) = \frac{a}{b}\tan(t)$, so $\cos(t) = \frac{a}{\sqrt{a^2 + b^2\tan^2(\theta)}}$, $\sin(t) = \frac{b \tan(\theta)}{\sqrt{a^2 + b^2 \tan^2(\theta)}}$. Hence the foot of the perpendicular is \begin{align*} \frac{1}{\sqrt{a^2 + b^2\tan^2(\theta)}}x + \frac{\tan^2(\theta)}{\sqrt{a^2 + b^2\tan^2(\theta)}}(x - x_0) &= 1\\ \frac{\sec^2(\theta) x}{\sqrt{a^2 + b^2\tan^2(\theta)}} &= 1 + \frac{\tan^2(\theta) x_0}{\sqrt{a^2 + b^2\tan^2(\theta)}}\\ x &= \cos(\theta)\sqrt{a^2\cos^2(\theta) + b^2\sin^2(\theta)} + \sin^2(\theta)x_0\\ y &= \sin(\theta)\sqrt{a^2\cos^2(\theta) + b^2\sin^2(\theta)} - \sin(\theta)\cos(\theta)x_0 \end{align*} If we shift the origin to $X$, we have the foot as $$\left ( \cos(\theta)\left (\sqrt{a^2\cos^2(\theta) + b^2\sin^2(\theta)} - \cos(\theta)x_0 \right ), \sin(\theta) \left (\sqrt{a^2\cos^2(\theta) + b^2\sin^2(\theta)} - \cos(\theta)x_0 \right ) \right )$$ Shifting to polar form, we have $$r = \sqrt{a^2\cos^2(\theta) + b^2\sin^2(\theta)} - \cos(\theta)x_0$$ The area contained in this is \begin{align*} A &= \int_0^{2\pi} \frac{1}{2}r^2 \ \mathrm{d}\theta\\ &= \frac{1}{2}\int_0^{2\pi} (a^2 + x_0^2) \cos^2(\theta) + b^2 \sin^2(\theta) - 2\cos(\theta)x_0\sqrt{a^2\cos^2(\theta) + b^2\sin^2(\theta)} \ \mathrm{d}\theta\\ &= \frac{1}{2}\pi(a^2 + b^2 + x_0^2) \end{align*} Here, we note that the last term of the integral evaluates to $0$ in the domains $[0, \pi]$ and $[\pi, 2\pi]$ so it vanishes.
Thus, the area enclosed is $$\frac{1}{2}\pi(a^2 + b^2 + x_0^2).$$
This area is minimised when $x_0 = 0$, so the origin minimises the area of the pedal curve of an ellipse.