Triangles are formed by pair of tangents drawn from any point on ellipse. Prove that the orthocentre of each triangle lies on the ellipse

Question

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

Answer 1

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

$$D=\begin{pmatrix}1&0&0\\0&1&0\\0&0&0\end{pmatrix}$$

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.

Answer 2

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.