Common Conjugate Diameters of Conics

Question

Find the equation to the common conjugate diameters of the conics $x^2+4xy+6y^2=1$ and $6x^2+6xy+9y^2=1$.

I dont know how to start here.

Answer 1

These equations are invariant by transformation $x \to -x, y \to -y$. As can be seen on figure below they represent ellipses centered at the origin.

Thus, we can assume that at least one of the common conjuguate diameters has equation $y=ax$ (see remark below).

Consequently, parallel lines to this diameter have equations:

\begin{equation} y=ax+b \end{equation}

In order to know the abscissas of the intersection of these parallels with each ellipse, we plug the above expression of $y$ into their equations, giving resp.

\begin{equation}x^2+4x(ax+b)+6(ax+b)^2=1 \ \iff (6a^2+4a+1)x^2+4b(3a+1)x+(6b^2-1)=0\end{equation}

\begin{equation}2x^2+2x(ax+b)+3(ax+b)^2=\frac13 \ \iff \ (3a^2+2a+2)x^2+2b(3a+1)x+(3b^2-\frac13)=0\end{equation}

Taking the half sum of the roots in each equation (classical formula : the sum of the roots of a quadratic equation $Ax^2+Bx+C=0$ is $-B/A$), we obtain the midpoints' abscissas. As these abscissas must coincide for both ellipses, we must have :

\begin{equation}\dfrac{2(3a+1)b}{6a^2+4a+1}=\dfrac{(3a+1)b}{3a^2+2a+2}\end{equation}

(for all $b$ within a certain range) giving evident solution

\begin{equation}a=-1/3\end{equation}

The first diameter has thus equation $y=ax=-x/3$.

The set of midpoints is constituted by points

\begin{equation} x=\dfrac{2(3a+1)b}{6a^2+4a+1} , \ \ y=ax+b \end{equation}

which is plainly (don't forget that $a=-1/3$) the set of points $(x=0,y=b)$, i.e. the vertical axis (see figure).

Remark : form $y=ax$ excludes vertical lines. But there is no loss of generality in our solution : if there is a vertical diameter (we have seen that this can happen...), we switch to its conjugate diameter which cannot be vertical too.

Important remark : If you are acquainted with linear algebra, this issue (finding conjugate diameters) is identical to the classical question of simultaneous diagonalization of quadratic forms. In both cases, you have to ''skew'' axes (the ''skewing'' being achieved by a matrix $S$) in order that one of the conics looks, with respect to these new axes (and with a certain imagination...) as a circle and the other as an ellipse with the new axes as its proper axes. This can be seen in the following way. Let $A$ and $B$ be the matrices associated to the LHS of the equations of these conic sections, and $S$ the linear transform associated to the ''skewing'' effect, i.e., a change of axes (former $x$ axis being transformed into the axis directed by vector $\binom{\ 3}{-1}$ ; ordinate axis unchanged):

\begin{equation}A=\begin{pmatrix}1 & 2\\ 2 & 6\end{pmatrix}, \ \ \ \ \ \ B=\begin{pmatrix}6 & 3\\ 3 & 9\end{pmatrix}, \ \ \ \ \ \ S=\begin{pmatrix}\ \ 3 & 0\\ -1 & 1\end{pmatrix}. \end{equation}

Then $A$ and $B$ become diagonalized, with respect to this new basis, under the form:

\begin{equation} A_1=S^TAS=\begin{pmatrix}1 & 0\\ 0 & 1\end{pmatrix}=I, \ \ \ \ \ \ \ \ \ B_1=S^TBS=\begin{pmatrix}15 & 0\\ 0 & \tfrac32\end{pmatrix}. \end{equation}

The square roots of the coefficients found in the diagonal of $B$: $\sqrt{15}$ and $\sqrt{\tfrac32}$ are the enlargment/shrinking factors to be used when one must pass from an ellipse to the other one.

Here we have only checked that the issue was the same (we had found previously the coefficients of $S$). But one can build an efficient method to obtain directly the coefficients of $S$ with matrix computations only. See for example the first solution to this question:

A property of positive definite matrices