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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.

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    $\begingroup$ Use sosmath.com/CBB/viewtopic.php?t=17029 $\endgroup$ – lab bhattacharjee Apr 15 '17 at 13:52
  • $\begingroup$ You haven’t given an equation for the conic, so at this point it’s impossible to find the axis lengths. Presumably, the expression you’ve written above is equal to zero. $\endgroup$ – amd Apr 15 '17 at 19:50
  • $\begingroup$ Please show your work up to the point that you get stuck so that we have a better idea of how to help you. $\endgroup$ – amd Apr 15 '17 at 19:55
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    $\begingroup$ you might find what you need in this book books.google.com/… $\endgroup$ – Will Jagy Apr 15 '17 at 21:31
  • $\begingroup$ lab bhattacharjee So basically i have to rotate the curve to remove xy term and find the equations of major and minor axis and then again rotate the axis to its initial state.. i guess that works.. $\endgroup$ – Abhash Jha Apr 16 '17 at 1:17
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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.

the ellipse

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  • $\begingroup$ sorry im not familiar with eigen values ..im still in high school ..still thanks i already got answer..and yeah ur final answer is wrong..plz check $\endgroup$ – Abhash Jha Apr 18 '17 at 8:44
  • $\begingroup$ @AbhashJha: Which answer is wrong? $\endgroup$ – Jan-Magnus Økland Apr 18 '17 at 8:46
  • $\begingroup$ the equation of axes.. $\endgroup$ – Abhash Jha Apr 18 '17 at 8:47
  • $\begingroup$ @AbhashJha: The fit on the figure is good. $\endgroup$ – Jan-Magnus Økland Apr 18 '17 at 8:48
  • $\begingroup$ i dont know .. by the way this question was asked in a mcq pattern and none of the option match your answer $\endgroup$ – Abhash Jha Apr 18 '17 at 8:58

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