Eccentricity of the ellipse $ax^2+2hxy+by^2=1$

Question

Find the eccentricity of the ellipse $ax^2+2hxy+by^2=1$

Solution:

The ellipse can be transformed to the principal axis as $\frac{x'^2}{\alpha^2}+\frac{y'^2}{\beta^2}=1$, with associated eccentricity, $e^2=\frac{\alpha^2-\beta^2}{\alpha^2}$

$\frac1{\alpha^2}+\frac1{\beta^2}=a+b, \frac1{\alpha^2\beta^2}=ab-h^2$

After that, $\alpha^2-\beta^2$ can be found in terms of $a,b,h$ and the expression of eccentricity can be found.

I wonder how did the following expressions come by.

$\frac1{\alpha^2}+\frac1{\beta^2}=a+b, \frac1{\alpha^2\beta^2}=ab-h^2$

Answer 1

The equations $ax^2+2hxy+by^2=1$ and $\frac{x'^2}{\alpha^2}+\frac{y'^2}{\beta^2}=1$ can be rewritten in matrix notation as $$ \begin{pmatrix}x & y\end{pmatrix} \begin{pmatrix}a & h\\ h & b\end{pmatrix} \begin{pmatrix}x \\ y\end{pmatrix} = 1, \tag{1} $$ $$ \begin{pmatrix}x' & y'\end{pmatrix} \begin{pmatrix}\alpha^{-2} & 0\\ 0 & \beta^{-2}\end{pmatrix} \begin{pmatrix}x' \\ y'\end{pmatrix} = 1. \tag{2} $$ Since $(x',y')$ is obtained from $(x,y)$ by a rotation $R(\theta)$, $$ \begin{pmatrix}x' \\ y'\end{pmatrix}=R(\theta)\begin{pmatrix}x \\ y\end{pmatrix}, \tag{3} $$ the matrices in $(1)$ and $(2)$ are related by $$ \begin{pmatrix}a & h\\ h & b\end{pmatrix} =R(\theta)^T\begin{pmatrix}\alpha^{-2} & 0\\ 0 & \beta^{-2}\end{pmatrix}R(\theta). \tag{4} $$ And since the rotation matrix $R(\theta)$ is orthogonal, we have the following identities: $$ \text{tr}\begin{pmatrix}a & h\\ h & b\end{pmatrix} =\text{tr}\begin{pmatrix}\alpha^{-2} & 0\\ 0 & \beta^{-2}\end{pmatrix} \implies a+b=\alpha^{-2}+\beta^{-2} \tag{5} $$ and $$ \det\begin{pmatrix}a & h\\ h & b\end{pmatrix} =\det\begin{pmatrix}\alpha^{-2} & 0\\ 0 & \beta^{-2}\end{pmatrix} \implies ab-h^2=\alpha^{-2}\beta^{-2}. \tag{6} $$

Answer 2

The centre of the ellipse can be found by partially differentiating the LHS w.r.t $x$ and $y$ and equating both the expressions obtained to $0$. This is a consequence of Taylor’s series for a function in $2$ variables. So, we get the system of equations

$$2ax+2hy=0$$ $$2hx+2by=0$$

Clearly, the origin is the centre of this ellipse. So, we can assume the coordinates of any point on this ellipse as $(r\cos\theta, r\sin\theta)$, i.e., in polar coordinate form. Then, on expressing $r$ as a function in $\theta$, we can get its maximum and minimum values which will correspond to the lengths of the semi-major(say $\alpha$) and teh semi-minor axes(say $\beta$) respectively. Doing so, we get:

$$\frac1{r^2}=\frac{a+b}2+\frac{a-b}2\cos2\theta+h\sin2\theta$$

Since the range of $a\cos\theta+b\sin\theta$ is $\left[-\sqrt{a^2+b^2}, \sqrt{a^2+b^2}\right]$, we get

$$\frac1{\alpha^2}=\frac{a+b}2+\sqrt{\left(\frac{a-b}2\right)^2+h^2}$$ $$\frac1{\beta^2}=\frac{a+b}2-\sqrt{\left(\frac{a-b}2\right)^2+h^2}$$

which gives $$\frac1{\alpha^2}+\frac1{\beta^2}=a+b$$ $$\frac1{\alpha^2\beta^2}=ab-h^2$$

Answer 3

Alternate approach

The equation can be written in matrix form as $$\mathbf{x}^T A \mathbf{x} = 1$$ where $$\mathbf{x} = \begin{bmatrix} x \\ y \end{bmatrix} \quad \text{and} \quad A = \begin{bmatrix} a & h \\ h & b \end{bmatrix}$$ Now, we have to diagonalize the matrix $A$

To diagonalize $A$, we find its eigenvalues $\lambda_1$ and $\lambda_2$ by solving the characteristic equation $\det(A - \lambda I) = 0$, which expands to $$\det \begin{bmatrix} a - \lambda & h \\ h & b - \lambda \end{bmatrix} = 0$$ This simplifies to $$(a - \lambda)(b - \lambda) - h^2 = 0$$ or $$\lambda^2 - (a + b)\lambda + (ab - h^2) = 0$$ Solving this quadratic equation gives the eigenvalues $\lambda_1 $ and $ \lambda_2$

The semi-major axis $ a' $ and semi-minor axis $b'$ are related to the eigenvalues by $$a' = \frac{1}{\sqrt{\lambda_{\text{min}}}}, \quad b' = \frac{1}{\sqrt{\lambda_{\text{max}}}}$$ where $ \lambda_{\text{min}} $ and $ \lambda_{\text{max}} $ are the smaller and larger eigenvalues, respectively.

The eccentricity $e $ of the ellipse is given by $$e = \sqrt{1 - \left(\frac{b'}{a'}\right)^2}$$ $\implies$ $$e = \sqrt{1 - \frac{\lambda_{\text{min}}}{\lambda_{\text{max}}}}$$

Answer 4

Hint:

Consider the general form $$ AX^2+2BXY+CY^2.\tag1 $$ Under the rotation transformation: $X=x\cos\theta+y\sin\theta$, $Y=-x\sin\theta+y\cos\theta$ the equation (1) takes on the form: $$\begin{align} x^2&\left[\frac{A+C}2+\frac{A-C}2\cos2\theta-B\sin2\theta\right]\tag2\\ +xy&\Big[(A-C)\sin2\theta+2B\cos2\theta\Big]\tag3\\ +y^2&\left[\frac{A+C}2-\frac{A-C}2\cos2\theta+B\sin2\theta\right]\tag4 \end{align}$$

Observe that the sum of coefficients at $x^2$ and $y^2$ is $A+C$. This holds always, in particular when the coefficient at $xy$ is $0$ for the appropiate value of $\theta$. This value of $\theta$ will be required while computing the product of the coefficients at $x^2$ and $y^2$.