# 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

• You refer to triangle $PBC$ - is that supposed to be $PAB$? $C$ hasn't been defined. – jmerry Mar 5 at 9:33
• Indeed, I fixed that, thanks – Daniel Monroe Mar 5 at 9:39

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.