4
$\begingroup$

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

$\endgroup$
  • $\begingroup$ You refer to triangle $PBC$ - is that supposed to be $PAB$? $C$ hasn't been defined. $\endgroup$ – jmerry Mar 5 at 9:33
  • $\begingroup$ Indeed, I fixed that, thanks $\endgroup$ – Daniel Monroe Mar 5 at 9:39
3
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.