Breaking down the equation of conic section

This is a second degree conic section in 2 variables . It's an equation of a eclipse .In this question we are asked to find the centre , eccentricity and the length of axes. I was able to get the centre of the conic by partial differentiating the question w.r.t x and y and solving them .But I was unable to find the eccentricity and the length of the axes . Is there anyway in which we can find this ?

• How? Could you explain how you got it Jan 17, 2022 at 13:38
• Sorry my remark was not correct. Jan 17, 2022 at 14:19
– MvG
Jan 17, 2022 at 15:50
• Not eclipse, but ellipse... Jan 17, 2022 at 21:01

The standard answer is that of rotating and translating the ellipse so that its axes coincide with cartesian axes. But if you are only interested in finding center and axes of the ellipse, I'll show you an alternate approach, which might spare you some computing.

Let $$f=9 x^2 + 4 xy + 6 y^2 - 22 x - 16 y + 9.$$ The first step is that of finding the points $$A$$, $$B$$ of the ellipse where the tangent is parallel to $$x$$ axis. This can be achieved by solving simultaneously $$f=0$$ and $$\partial f/\partial x=0$$, with the result: $$A,B=\left(1\pm{2\sqrt5\over15},1\pm{3\sqrt5\over5}\right).$$ From that we get the center $$O=(1,1)$$, which is the midpoint of $$AB$$. And $$AB$$ is a diameter of the ellipse, which is the conjugate of the diameter $$CD$$ parallel to $$x$$ axis. Points $$C$$, $$D$$ can be found by solving simultaneously $$f=0$$ and $$y=1$$, with the result: $$C,D=\left(1\pm{\sqrt{10}\over3},1\right).$$ But semi-axes $$a$$, $$b$$ of the ellipse are related to semi-diameters $$OA$$, $$OC$$ by Apollonius' relations: $$a^2+b^2=OA^2+OC^2,\quad ab=2\cdot Area(AOC).$$ This leads to the system of equations: $$\cases{ a^2+b^2=3\\ ab=\sqrt2 }$$ which can be easily solved to $$a=\sqrt2$$, $$b=1$$.

By cancelling the gradient of $$f$$, we find the center $$(1,1)$$. And by translation the centered conic:

$$9u^2+4uv+6v^2-10=0.$$

We can find the axis length by minimization/maximization of the squared distance to the origin under the above constraint, using Lagrange multipliers.

We cancel the gradient of $$u^2+v^2+\lambda(9u^2+4uv+6v^2-10),$$

giving

$$\begin{cases}2u+\lambda(18u+4v)=0,\\2v+\lambda(4u+12v)=0.\end{cases}$$ and by elimination of $$\lambda$$,

$$2u^2-3uv-2v^2=0=(2u+v)(u-2v)$$ which gives the equations of the axes.

Finally, by eliminating $$u$$ or $$v$$, and solving quadratic equations, you can get the endpoints of the axes, their lengths and the eccentricity.