# Ellipse in Quadratic Form: Finding Intercepts with Principal Axes

Where an ellipse is expressed in quadratic form (e.g. $ax^2 + bxy + cy^2 = k$ is expressed as $x^TQx = k$), the principal axes are in the directions given by the eigenvectors of Q. I understand this.

Now, if we want to find the intercepts of the ellipse and the principal axes, apparently they can be found as intercepts = $\pm \sqrt{k/\lambda_i}$ (where $\lambda_i$ is the eigenvalue corresponding to the eigenvector which describes the axis we are looking for the intercept with).

What I don't understand is how to derive this. I can see how to derive the intercepts in terms of $a, b, c, k$ but not in terms of the eigenvaluesand $k$.

Can someone please help shed some light on this - perhaps provide a derivation?

Thanks in advance for your help.

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I think you mean $ax^2+bxy+cy^2=k$ – Brett Frankel Apr 30 '12 at 2:59
@BrettFrankel sorry - that was a typo. Thanks for letting me know! – JonaGik Apr 30 '12 at 4:02

## 1 Answer

Suppose $(\hat x , \hat y)$ is an eigenvector corresponding to the eigenvalue $\lambda_i$. In this case $Q$ is given by: $$Q = \left( \begin{array}{cc} a & \frac{1}{2} b \\ \frac{1}{2} b & c \end{array} \right).$$ Then we have $Q \binom{\hat x}{\hat y} = \lambda_i \binom{\hat x}{\hat y}$, and so $\binom{\hat x}{\hat y}^T Q \binom{\hat x}{\hat y} = a\hat x^2 + b\hat x \hat y + c \hat y^2 = \lambda_i (\hat x^2 + \hat y^2)$. To find out where the ellipse intersects the principal axis (ie, the line going through $0$ and $(\hat x, \hat y)$), we need to scale the eigenvector so it lies on the ellipse. In other words, we need to choose a $\sigma$ such that $\binom{\sigma \hat x}{\sigma \hat y}^T Q \binom{\sigma \hat x}{\sigma \hat y} = \lambda_i ((\sigma \hat x)^2 + (\sigma \hat y)^2)$ = k. Thus the distance from the origin to the intercept is given by: $$\sqrt{(\sigma \hat x)^2 + (\sigma \hat y)^2} = \sqrt{\frac{k}{\lambda_i}}$$

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Thank you very much! – JonaGik Apr 30 '12 at 4:31