A triangle is formed by pair of tangents drawn from a point on the ellipse $$a^2x^2+b^2y^2=(a^2+b^2)^2$$ to the ellipse $$b^2x^2+a^2y^2=a^2b^2$$ and their chord of contact.

Find the locus of orthocentre of triangle

  • $\begingroup$ Do you know anything about pole/polar relationships? The chord of contact is the polar line of the point from which you’re drawing the tangents, so that seems like it could be of use. $\endgroup$
    – amd
    Nov 1 '17 at 19:52
  • $\begingroup$ Perhaps not. It was an obvious thing that jumped out at me. I can get this result by working “backwards”—starting from an arbitrary point and deriving the outer ellipse from the constraint that the orthocenter lie on the inner ellipse, but it’s not pretty. $\endgroup$
    – amd
    Nov 2 '17 at 2:31
  • $\begingroup$ Just to clarify, on which ellipse does the orthocentre lie on? $\endgroup$
    – Dylan
    Nov 2 '17 at 4:04
  • $\begingroup$ So we don't know which ellipse it is, just to prove that it is an ellipse? I'm asking because the problem is not clear $\endgroup$
    – Dylan
    Nov 2 '17 at 15:08
  • 2
    $\begingroup$ Now I am confused. To me the original question was kind of obvious: “the ellipse” referring to the one to which the tangents where drawn, i.e. the second one. This is a much stronger (and still true) statement than what we have now, namely only showing that the locus is any ellipse. I would revert the edit except for OP's comment just now that any ellipse will do. So I wonder how certain OP is about the goal of this question. $\endgroup$
    – MvG
    Nov 2 '17 at 21:27

One way to do this is with coordinates and a bit of computer algebra. Pick a generic point on the first ellipse using the tangent half-angle formula:

$$x=\frac{(1-t^2)(a^2+b^2)}{(1+t^2)a}\qquad y=\frac{(2t)(a^2+b^2)}{(1+t^2)b}$$

Or in homogeneous coordinates:

$$p_1= \begin{pmatrix}(1-t^2)(a^2+b^2)b\\(2t)(a^2+b^2)a\\(1+t^2)ab\end{pmatrix}$$

Also write your ellipses as matrices for use with homogeneous coordinates:

$$E_1=\begin{pmatrix}a^2&0&0\\0&b^2&0\\0&0&-(a^2+b^2)^2\end{pmatrix}\qquad E_2=\begin{pmatrix}b^2&0&0\\0&a^2&0\\0&0&-a^2b^2\end{pmatrix}$$

You can verify that $p_1^T\cdot E_1\cdot p_1=0$, so $p_1$ lies on $E_1$ as intended. Now the polar of $p_1$ with respect to $E_2$ is $g_1=E_2\cdot p_1$.

Next we need to intersect that line $g_1$ with the conic $E_2$, following this approach for a conic-line intersection. Write $\hat g_1$ for the cross product matrix of $g_1$. Then

$$M_1 = \hat g_1^T\cdot E_2\cdot\hat g_1$$

is a symmetric matrix of rank $2$ which encodes the two points of intersection. To access them, look at

$$M_2 = M_1 + \mu\hat g_1$$

which will be of rank $1$ for two specific values of $\mu$. To find them, compute the determinant of any $2\times2$ submatrix and you will find the solutions to be

$$\mu=\pm a^2b^2\sqrt{2 a^{2} b^{6} t^{4} + b^{8} t^{4} + 4 a^{8} t^{2} + 8 a^{6} b^{2} t^{2} - 4 a^{2} b^{6} t^{2} - 2 b^{8} t^{2} + 2 a^{2} b^{6} + b^{8}}$$

This is where things get ugly, as $\mu$ is in general not rational even for a rational choice of $a,b,t$. You can use $\mu$ as a symbolic variable for the time being, though. As $M_2$ is of rank $1$, all rows are multiples of one another, so they are homogeneous coordinates of the same point. Likewise all columns. Let $p_2$ be the first row and $p_3$ be the first column. These are your points of intersection, in general not rational.

The corresponding tangent lines can be computed as

$$g_2 = E_2\cdot p_3\qquad g_3 = E_2\cdot p_2$$

You could also compute $g_2=p_1\times p_3$ but that leads to more complicated expressions representing the same lines. You can use the matrix


to turn any line into the point at infinity orthogonal to it. So $D\cdot g_2$ is the point through which all lines orthogonal to $g_2$ pass. Thus

$$q = \bigl((D\cdot g_2)\times p_2\bigr)\times\bigl((D\cdot g_3)\times p_3\bigr)$$

is the orthocenter, computed as the intersection of the two lines orthogonal to the two sides and passing through the opposite triangle corners.

This point $q$ will lie on the ellipse $E_2$ if $q^T\cdot E_2\cdot q=0$. Factorizing this, my computer algebra system tells me that

\begin{align*} q^T\cdot E_2\cdot q &= 16 \cdot t^{2} \cdot (t - 1)^{2} \cdot (t + 1)^{2} \cdot (t^{2} + 1)^{2} \cdot (b^{2} t^{2} - 2 a^{2} t - 2 b^{2} t + b^{2})^{2} \\&\quad\cdot (b^{2} t^{2} + 2 a^{2} t + 2 b^{2} t + b^{2})^{2} \cdot (a^{2} + b^{2})^{4} \cdot b^{16} \cdot a^{40} \cdot \mu^{2} \\&\quad\cdot (\mu^{2} - 2 a^{6} b^{10} t^{4} - a^{4} b^{12} t^{4} - 4 a^{12} b^{4} t^{2} - 8 a^{10} b^{6} t^{2} + 4 a^{6} b^{10} t^{2}\\&\qquad + 2 a^{4} b^{12} t^{2} - 2 a^{6} b^{10} - a^{4} b^{12})^{2} \end{align*}

As you can see, this is of degree $6$ in $\mu$, but only the even powers of $\mu$ occur. So substituting

$$\mu^2=b^{4} \cdot a^{4} \cdot (2 a^{2} b^{6} t^{4} + b^{8} t^{4} + 4 a^{8} t^{2} + 8 a^{6} b^{2} t^{2} - 4 a^{2} b^{6} t^{2} - 2 b^{8} t^{2} + 2 a^{2} b^{6} + b^{8})$$

you can evaluate this term (or just the last big factor) and find that is indeed identical zero. Q.e.d.

With a bit more effort, you can also strip common factors from $q$, taking the substitution for $\mu^2$ into account, and find a much simpler representative

$$q\sim q_2=\begin{pmatrix}(1-t^2)a\\2tb\\(1+t^2)\end{pmatrix}$$

which satisfies $q_2^T\cdot E_2\cdot q_2=0$ in a way you can verify by hand.

It is noteworthy that this result is how you'd express a generic point on $E_2$ using the tangent half-angle formula. So the whole map from $p_1$ to $q$ is essentially what you get if you rescale the $x$ and $y$ direction independently.

$$x'=\frac{(1-t^2)a}{1+t^2}=\frac{a^2}{a^2+b^2}x\qquad y'=\frac{(2t)b}{1+t^2}=\frac{b^2}{a^2+b^2}y$$

Someone might be able to build a nicer proof from this.

  • $\begingroup$ Can't think of any obvious simplifications. Piquito's answer is simpler in several ways. Personally I consider my approach pretty simple to work with, since it's “only polynomials”, but I concede that it's based on several years of working with projective geometry to establish the concept's I'm using. The links I included can provide some of that, ISBN 3642172857 for more. If you have any specific questions, you might ask about them in a comment or perhaps even a separate question, but I doubt that me rewriting the answer to use different words would make it much simpler. $\endgroup$
    – MvG
    Nov 2 '17 at 21:33
  • $\begingroup$ I followed this process, too, but starting with an arbitrary point $(\xi,\eta)$ external to the inner ellipse. If you then substitute $(a^2+b^2)/a\cos t$ and $(a^2+b^2)/b\sin t$ for $\xi$ and $\eta$, respectively, once you’ve computed the orthocenter $q$ this way, it all eventually simplifies down to $q=(a\cos t,b\sin t)$. I was surprised that the parameterizations synced up like that. An ugly computation either way. Seems like there should be some more clever approach that I’m missing than simply grinding it out. I’ll think about your last observation and see where it leads. $\endgroup$
    – amd
    Nov 2 '17 at 22:19

HINT.-MvG's answer tells you that the calculations can be long and apparently complicated. I give you here a mode that may seem shorter and easier for you.

Let $E_1$ and $E_2$ the two ellipses so$$E_1:\frac{x^2}{A^2}+\frac{y^2}{B^2}=1;\space A=\frac{a^2+b^2}{a},\space B=\frac{a^2+b^2}{b}\\E_2: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$ It is convenient to start from two points $P(x_1,y_1)$ and $Q(x_2,y_2)$ of $E_2$. The corresponding tangents are $$T_1:\frac{x_1X}{a^2}+\frac{y_1Y}{b^2}=1\text{ with pente }-\frac{x_1b^2}{y_1a^2}\\T_2:\frac{x_2X}{a^2}+\frac{y_2Y}{b^2}=1\text{ with pente }-\frac{x_2b^2}{y_2a^2}$$ $T_1$ and $T_2$ go through the point $T(x_0,y_0)\in E_1$ so we have $$\frac{x_1x_0}{a^2}+\frac{y_1y_0}{b^2}=\frac{x_2x_0}{a^2}+\frac{y_2y_0}{b^2}\iff x_0\frac{x_1-x_2}{a^2}=y_0\frac{y_2-y_1}{b^2}\space (*)$$ The perpendicular from $P(x_1,y_1)$ to the line $QT$ has equation $$(Y-y_1)=(X-x_1)\frac{y_2a^2}{x_2b^2}$$ similarly we have

$$(Y-y_2)=(X-x_2)\frac{y_1a^2}{x_1b^2}$$ These two perpendiculars intersect at the orthocentre $H$ of the post and this common point should belong to the elipse $E_2$ which can be verified using the relation $(*)$ above and the coordinates of the found orthocentre (keep in mind that $\frac{x_0^2}{A^2}+\frac{y_0^2}{B^2}=1$).

I leave to you the (non hard) final calculation.

  • $\begingroup$ “$T_1$ and $T_2$ go through the point $T(x_0,y_0)\in E_1$...” Since $T_1$ and $T_2$ are an arbitrary pair of tangents to $E_2$, how do you know that their intersection lies on $E_1$? They might not even intersect in the first place. $\endgroup$
    – amd
    Nov 2 '17 at 21:52
  • $\begingroup$ Not so, the expression (*) forces point $T$ in such a way that the conditions of the problem are met. (Please believe me, I was expecting a reply similar to yours but from Abc Def). $\endgroup$
    – Piquito
    Nov 2 '17 at 23:11
  • $\begingroup$ @amd: I understand that your reply is due to my bad English. What I meant is that if we want the tangents to leave the point T then we put the condition (*). Precisely because the tangents are arbitrary I can do with it what I want. $\endgroup$
    – Piquito
    Nov 2 '17 at 23:28

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