Locus of orthocenter of triangle inscribed in ellipse While messing around with ellipses in Geogebra, I found the following interesting result:
Let $\alpha$ be an ellipse. Let $AB$ be a fixed chord, and let $P$ be a point that moves freely on $\alpha$. Then as $P$ traces the ellipse, the orthocenter of triangle $PAB$ traces another ellipse(which passes through $AB$).
If you take the $\alpha$ to be of the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, then interestingly the foci of the ellipse drawn by the orthocenter is parallel to the $y$-axis. The two ellipses also seem to have the same eccentricity. 
I believe a proof could be along the lines of applying the linear transformation given by the matrix $\begin{bmatrix} \frac{1}{a} && 1 \\ 1 && \frac{1}{b} \end{bmatrix}$ which takes the ellipse to the unit circle, solving the locus in the circle case (which turns out to be just the unit circle reflected after $AB$) and then applying the reverse transformation, but it does not seem to work very well for some reason.
I also thought about parametrizing the points and using complex numbers; I obtained a formula for the orthocenter in terms of an angle $\phi$ (where $P = (a\cos(\phi), b\sin(\phi))$ but it seems very unwieldly to show that the points form an ellipse, especially since I haven't been able to guess the real part of the foci.
Any ideas would be welcome.
image
 A: With the help of Mathematica, I was able to confirm your suspicion. 

Let the ellipse be parameterized by $(a \cos\theta, b \sin\theta)$, and let $A$ and $B$ be points corresponding to $\theta = 2\alpha$ and $\theta = 2\beta$. After some symbol-crunching, we find that the orthocenter $(x,y)$ satisfies
$$\begin{align}
ax &= \left(\; a^2\sin^2(\alpha+\beta) + b^2 \cos^2(\alpha + \beta) \;\right)\cos\theta \\[4pt]
&+\left(a^2-b^2\right)\cos(\alpha+\beta)\sin(\alpha+\beta) \sin\theta \\[4pt]
&+\left(a^2 + b^2\right) \cos(\alpha-\beta)\cos(\alpha+\beta) \\[18pt]
-by&=\left(a^2 - b^2\right) \cos(\alpha+\beta) \sin(\alpha+\beta) \cos\theta \\[4pt]
&- \left(\; a^2 \sin^2(\alpha+\beta)+ b^2 \cos^2(\alpha + \beta) \;\right)\sin\theta \\[4pt]
&- \left(a^2 + b^2\right) \cos(\alpha-\beta) \sin(\alpha+\beta)
\end{align} \tag{1}$$
Solving system $(1)$ for $\cos\theta$ and $\sin\theta$, substituting into $\cos^2\theta+\sin^2\theta=1$, and simplifying, we obtain the equation of a new ellipse:
$$\begin{align}
&\phantom{+\;\;}a^2 \left(x - \frac{a^2 + b^2}{a} \cos(\alpha-\beta) \cos(\alpha+\beta)\right)^2 \\[4pt]
&+b^2 \left(y - \frac{a^2 + b^2}{b} \cos(\alpha-\beta) \sin(\alpha+\beta)\right)^2 \\[4pt]
&=a^4 \sin^2(\alpha+\beta) + b^4 \cos^2(\alpha+\beta)
\end{align} \tag{$\star$}$$
Since the horizontal and vertical radii are proportional to $1/a$ and $1/b$, we see that this ellipse's major and minor axes are perpendicular to the original's axes, respectively. Moreover, for $a\geq b$, the eccentricity is $\sqrt{a^2-b^2}/a$, which matches that of the original ellipse.

More generally, we can take a conic of eccentricity $e$, parameterized by 
$$\frac{p}{1+e \cos\theta}\left(\cos\theta,\sin\theta\right) $$
whose focus is at the origin and whose major/transverse axis coincides with the $x$-axis. Taking $A$ and $B$ to correspond to $\theta=\alpha$ and $\theta = \beta$, we can perform the same kind of analysis as above to obtain a comparable conic:
$$\begin{align}
&\phantom{+\;\;}
   x^2 \phantom{\left( 1 - e^2 \right) }( 1 + e \cos\alpha ) ( 1 + e \cos\beta ) \\
&+ y^2 \left( 1 - e^2 \right) ( 1 + e \cos\alpha ) ( 1 + e \cos\beta ) \\
&- x p \left( \left( 2 + e^2 \right) \left( \cos\alpha + \cos\beta \right) + 2 e \left(1 + 2 \cos\alpha \cos\beta \right) \right) \\
&- y p \left( 2 - e^2 \right) ( \sin\alpha + \sin\beta + e \sin(\alpha+\beta) ) \\
= &- p^2 \left( 1 + 2\cos(\alpha-\beta) + e (\cos\alpha + \cos\beta ) - e^2 \sin\alpha \sin\beta \right)
\end{align}$$
For $e=1$, the original conic is a parabola; we see that the new conic is, too, since its $y^2$ term vanishes. Otherwise, the horizontal and vertical radii of the new conic are proportional to $1$ and $1/|1-e^2|$, respectively, and we deduce that the new eccentricity is also $e$. In all cases, we see that the new conic is rotated $90^\circ$ with respect to the original.
