I was trying to solve this question using vectors and ideas of gradient, all the context is same as original except the fact that point lines on the x-axis. I want here the point to lie anywhere on the plane.
Find the value of x0 at which the area of the pedal curve of the ellipse with respect to the point P( x0 , 0) is minimized
My attempt:
Let the tangent line be represented as $\tau$ and let the signed distance from tangent line to point be $d_{\tau}(p)$, finally let the unit normal vector of the tangent line be $n$.
Note that $p$ is given as an ordered pair of coordinates, in whatever coordinate system you use. Now, the foot of perpendicular is given as:
$$f_{\tau} (p) = p-d_{\tau} (p) n$$
Now, we can parameterize the unit normal as a function of $t$
$$ F_{\tau} (p,t) = f_{\tau} (p) = p - d_{\tau} \left( p\right) n(t)$$
Now, we want the area of Pedal curve, we can use the gauss shoelace formula for this, treating $F_{\tau}(t)$ as a position vector:
$$\begin{align} \int F_{\tau} (t,p) \times \frac{d}{dt} F_{\tau} (t,p) dt = \int \left[ p - d_{\tau}(p) n\right] \times \left[-d_{\tau}(p) \frac{dn}{dt}-\frac{d (d_{\tau}(p) )}{dt} n\right] dt=\\ \int \left[ d_{\tau}(p) \frac{dn}{dt}+n \frac{d (d_{\tau}(p) )}{dt} \right]\times \left[ p-d_{\tau}(p)n\right] dt= \int \frac{dn}{dt} \times \left[ p d_{\tau}(p)- (d_{\tau} (p))^2 n \right] + \frac{d (d_{\tau}(p) )}{dt}n \times \left[ p\right]dt \end{align}$$
The final problem is to find a point $p$ such that the integral:
$$I(p) = \int_{0}^{2\pi} \frac{dn}{dt} \times \left[ p d_{\tau} (p) - d_{\tau}(p)^2 n \right] + \frac{d (d_{\tau}(p) )}{dt}n \times \left[ p\right] dt$$
Is minimized.
Brute forcing by taking the parameterization $(a \cos \theta, b \sin \theta)$ of ellipse and the normals and $\frac{dn}{dt}$ calculations became literal hell.
I thought perhaps we could use the idea of the gradient and see where that vanishes to find max i.e: find where point $p$ such that:
$$ \nabla I(p) = 0$$
However, I am not sure how to compute the gradient of the integral in this case. Some help would be appreciated.
p.s: Ideally speaking I'd like to finish the proof completely with 0 use of coordinates.
1/8/2022: We can make a simplification for the integral:
$$G(p)= \oint \frac{dn}{dt} \times \left[ p d_{\tau} (p) - d_{\tau}(p)^2 n \right] dt \int_{0}^{2\pi} (\frac{dn}{dt} \times n) \left[ p_n d_{\tau} (p) - d_{\tau}(p)^2 \right] dt$$
Due to the fact that $ \frac{dn}{dt} \times p = \frac{dn}{dt} \times n d_{\tau}(p)$ where $d_{\tau}(p)$ is the component of $p$ along $n$ (i.e: vector of p to foot of perpendicular), hence $G(p)$ is zero everywhere.
Hence,
$$I(p)= \oint \frac{d (d_{\tau}(p) )}{dt}n \times \left[ p\right] dt$$
I suppose, I would have to nudge the above expression in some direction $ dp $ and check the first order variation:
$$I(p+ dp)= \oint n\frac{d}{dt} \left[ d_{\tau}(p) + \nabla d_{\tau}(p) \cdot dp \right] \times \left[ p+ dp\right]dt$$
Not sure how to continue.. $$ dI = \oint \left[ n \frac{d(d_{\tau} (p))}{dt} \times dp + n \left[ \frac{d}{dt} \nabla d_{\tau}(p) \cdot dp \right] \times p\right] +O(|dp|^2)dt$$
To repeat: The goal is to find what $p$ is
