Find major and minor axis of given conic.

Question

Find the equation of major and minor axis of the given conic.Also find length of major and minor axis.

$$5x^2 + 5y^2 +6xy + 22x - 26y + 29$$ =0

I tried finding the nature of conic and centre of conic by partially differentiating it but could not think ahead.

Answer 1

Rewrite the equation using matrices:

$\begin{pmatrix}x&y\end{pmatrix}\begin{pmatrix}5&3\\3&5\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}+\begin{pmatrix}22&-26\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}+29=0$

$M=\begin{pmatrix}5&3\\3&5\end{pmatrix}, K=\begin{pmatrix}22&-26\end{pmatrix}$

Diagonalising $M$ using the eigenvalues and eigenvectors we get:

$P=\begin{pmatrix}\frac{\sqrt{2}}{2}&-\frac{\sqrt{2}}{2}\\\frac{\sqrt{2}}{2}&\frac{\sqrt{2}}{2}\end{pmatrix}$

$\begin{pmatrix}x\\y\end{pmatrix}=P\begin{pmatrix}x'\\y'\end{pmatrix}, P^tMP=\begin{pmatrix}8&0\\0&2\end{pmatrix}$

$\begin{pmatrix}x'&y'\end{pmatrix}P^tMP\begin{pmatrix}x'\\y'\end{pmatrix}+KP\begin{pmatrix}x'\\y'\end{pmatrix}+29=0$

that is:

$8x'^2+2y'^2-2\sqrt{2}x'-24\sqrt{2}y'+29=0$

or completing the squares and rearranging:

$(\frac{x'-\frac{\sqrt{2}}{8}}{\frac18 \sqrt{922}})^2+(\frac{y'-6\sqrt{2}}{\frac14\sqrt{922}})^2=1$

Making the semi-major axis $\frac14\sqrt{922}$ and the semi-minor axis $\frac18 \sqrt{922}$.

Using the inverse transformation $\begin{pmatrix}x'\\y'\end{pmatrix}=P^t\begin{pmatrix}x\\y\end{pmatrix}:$

$x'-\frac{\sqrt{2}}{8}=0, \frac{\sqrt{2}}{2}x+\frac{\sqrt{2}}{2}y-\frac{\sqrt{2}}{8}=0$

we get the equation of one of the axes: $y=\frac14-x,$ and

$y'-6\sqrt{2}=0, -\frac{\sqrt{2}}{2}x+\frac{\sqrt{2}}{2}y-6\sqrt{2}=0$

the equation of the other axis $y=12+x.$

Edit:

In differentiating to get the center we get the following system:

$10x+6y+22=0$

$6x+10y-26=0$

Since by the comments we get a rotation by $\frac{\pi}{4}$ ($\tan(2\theta)=\frac{B}{A-C}$), let's try and find the two equations in this system that is rotated to that angle (adding and subtracting):

$16x+16y=4$

$-4x+4y=48$

These are the same two lines on the figure.

Answer 2

Find the center of the ellipse then shoft your origin to it then you will get an equation like this $$ax^2+2hxy+by^2=1$$ Now consider a circle $x^2+y^2=r^2$ a circle intersect ellipse in 4 points but if $r$ is semi major axis(or semi minor) the circle touches the ellipse in two points.

Now homogenize the equation using the circle $$ax^2+2hxy+by^2=(x^2+y^2)/r^2$$ Now write a quadratic and set the discriminant to 0.

You will get two values of $r$ giving you the length of semi major and semi minor axes.