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Let there be two ellipses centered at the origin, and rotated angles $\alpha$ and $\alpha'$ from the x-coordinate axis. $a, b$ are the semiaxes of the first, $a',b'$ the semiaxes of the second, and the ratio $ab/a'b'$ is called $f$. The equations for both ellipses can be written: $$ \frac{(x\cos\alpha+y\sin\alpha)^2}{a^2}+\frac{(y\cos\alpha-x\sin\alpha)^2}{b^2}=1 $$ and $$ \frac{(x\cos\alpha+fy\sin\alpha)^2}{a^2}+\frac{(fy\cos\alpha-x\sin\alpha)^2}{b^2}=1 $$ Where the equation for the second ellipse is written for convenience as a function of the parameters $a,b, \alpha$.

Now in the text I am following I find that the semiaxes of the second ellipse, $a', b'$, can be obtained from the second equation above, using the relation: $$ \frac{1}{a'^2}+\frac{1}{b'^2}=\frac{\cos^2\alpha+f^2\sin^2\alpha}{a^2}+\frac{\sin^2\alpha+f^2\cos^2\alpha}{b^2} $$ However I don't have any clue were this last relation comes from. Any help is welcome!

Andrei
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P Macmutton
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1 Answers1

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The equation of an ellipse in standard position with semi-axis lengths $a'$ and $b'$ can be written in matrix form as $$\begin{bmatrix}x&y\end{bmatrix}\begin{bmatrix}\frac1{a'^2}&0\\0&\frac1{b'^2}\end{bmatrix}\begin{bmatrix}x\\y\end{bmatrix}=1.\tag1$$ The trace of the central matrix—the sum of its main diagonal elements—in this equation is equal to $\frac1{a'^2}+\frac1{b'^2}$. Rotating this ellipse amounts to a similarity transformation of the coefficient matrix, which leaves its trace unchanged. Expanding the equation of the rotated ellipse given in the question and collecting terms gives $$\left({\cos^2\alpha\over a^2} + {\sin^2\alpha\over b^2}\right)x^2+2f\cos\alpha\sin\alpha\left(\frac1{a^2} - \frac1{b^2}\right)xy + \left({f^2\sin^2\alpha\over a^2} + {f^2\cos^2\alpha\over b^2} \right)y^2=1.\tag2$$ In matrix form, the diagonal elements are the coefficients of $x^2$ and $y^2$, so take these two coefficients from equation (2) and rearrange their sum to get $$\frac1{a'^2}+\frac1{b'^2}={\cos^2\alpha+f^2\sin^2\alpha\over a^2}+{\sin^2\alpha+f^2\cos^2\alpha\over b^2}$$ as required.

amd
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  • Thanks a lot, very instructive. Let me ask so it remains clear to me: using matrices you have shown me that the coefficients A and B of the general equation (e.g. equation 2) are the inverse of the square of the semiaxes ($1/a^2$ and $1/b^2$). Is this a rule that stands true for any ellipse? I mean is it always true that $A=1/a^2$ and $B=1/b^2$? It sounds awkward to me... – P Macmutton May 19 '17 at 12:11
  • Sorry I meant $C$ instead of $B$ (the coefficient of $y^2$). But my question stands. – P Macmutton May 19 '17 at 12:27
  • @PMacmutton No. What I’ve shown is that $A+C$ is invariant with respect to rotations. It’s also translation-invariant, since translations affect only the linear and constant terms of the equation. – amd May 19 '17 at 16:56
  • Ok... I understand that the equation of the ellipse in matrix form has $A$ and $C$ in the diagonal. What I don't know is where your Equation (1) comes from, or why it is equated with the general equation of the ellipse to get the required equation $1/a'^2 + 1/b'^2 = A + C$. – P Macmutton May 19 '17 at 18:17
  • @PMacmutton Multiply it out and you’ll see that it’s precisely the equation of an ellipse in standard position. For the ellipse $Ax^2+2Bxy+Cx^2=1$, the corresponding matrix is $\begin{bmatrix}A&B\B&C\end{bmatrix}$. If the matrix of an ellipse is $Q$, the matrix after rotation is $RQR^T$, where $R$ is the rotation matrix, but, as I wrote in my answer, this operation leaves the trace of the matrix—the sum of its diagonal elements—unchanged. – amd May 19 '17 at 20:36
  • Ok, ok, now I understand, the equality is valid for any ellipse in standard position and the same ellipse after rotation, where the trace is the same. Thank you very much @amd – P Macmutton May 19 '17 at 23:46