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Could someone show me how to find a polar form of this general equation of a conic section?

$x^2+y^2-xy+x=4$

I have managed to determine this is an ellipse and write it in a canonical form with changed variables:

${(x'+{1 \over \sqrt{2}})^2 \over (\sqrt{26 \over 3})^2}+{(y'-{1 \over 3\sqrt{2}})^2 \over (\sqrt{26 \over 9})^2}=1$

I'm wondering if I can use this form and take the value of the semi-major and the semi-minor axis, and then use the general polar form for ellipse:

$r(\phi)={a(1-e^2) \over 1+e*cos{\phi}}$

I see that the centre is different in the general and canonical form with changed variables. But, what about 'a' and 'b'? Are the values of them changed, too?

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  • $\begingroup$ Where is $ b $? $\endgroup$ – Nosrati May 15 at 7:28
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I am not sure that your canonical form is correct because you didn't define $x'$ and $y'$.

In order to bring the ellipse into standard position, one have to translate it and rotate it of $\pi /4$. This corresponds to the change of variables : $$\begin{cases} x=x_0+\frac{X-Y}{\sqrt{2}} \\ y=y_0+\frac{X+Y}{\sqrt{2}} \end{cases} \qquad\begin{cases} X=\frac{(y-y_0)+(x-x_0)}{\sqrt{2}} \\ Y=\frac{(y-y_0)-(x-x_0)}{\sqrt{2}} \end{cases}$$ Putting the new coordinates into the original equation leads to

$\left(x_0+\frac{X-Y}{\sqrt{2}} \right)^2+\left(y_0+\frac{X+Y}{\sqrt{2}} \right)^2 - \left(x_0+\frac{X-Y}{\sqrt{2}}\right) \left(y_0+\frac{X+Y}{\sqrt{2}} \right)+\left(x_0+\frac{X-Y}{\sqrt{2}}\right)=4$

After simplification the coefficients of $X$ and $Y$ are obtained. They must be nul for the canonical form in the new system of axes. $$\begin{cases} \frac{y_0+x_0+1}{\sqrt{2}}=0 \\ \frac{3y_0-3x_0-1}{\sqrt{2}}=0 \end{cases}\quad\implies\quad \begin{cases} x_0=-\frac23 \\ y_0=-\frac13 \end{cases}$$ $(x_0\:,\:y_0)$ are the coordinates of the origin of the new system of axes in the original system of axes. Don't confuse them with the coordinates of the origin of the original system of axes in the new system of axes.

The simplification of $\left(-\frac23 +\frac{X-Y}{\sqrt{2}} \right)^2+\left(-\frac13+\frac{X+Y}{\sqrt{2}} \right)^2 - \left( -\frac23 +\frac{X-Y}{\sqrt{2}}\right) \left( -\frac13 +\frac{X+Y}{\sqrt{2}} \right)+\left(-\frac23 +\frac{X-Y}{\sqrt{2}}\right)=4$ leads to : $$\boxed{\frac{3}{26} X^2+\frac{9}{26} Y^2=1}$$ This is the equation of the ellipse in the new system of axes.

Now you can use the usual formulas to compute the parameters of the ellipse (semiminor and semimajor axes, eccentricity,...) and the usual representation in polar coordinates (from the origin of the new system of axes).

One can come back to the original system of axes with :$\quad\begin{cases} X=\frac{(y+\frac13)+(x+\frac23)}{\sqrt{2}} \\ Y=\frac{(y+\frac13)-(x+\frac23)}{\sqrt{2}} \end{cases}$

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By standard rectangular to polar conversion

$$ x= r \cos \theta;\, y=r \sin \theta $$ we get $$r^2(1-\sin \theta \cos \theta)+ r \cos \theta-4= 0 $$

The quadratic can be solved for. You found required axis rotation. The given form does nor convert to Newton polar focus origin form you gave straight away.

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Polar coordinate system is not a good one to work on translation but is good for rotations of objects. The best you can do is to do as follows:

Substituting $x=r\cos\theta$ and $y=r\sin\theta$, we have $$r^2-r^2\cos\theta\sin\theta-r\cos\theta=4. $$ So, $$r=\frac{\cos\theta\pm\sqrt{\cos^2\theta+16-4\sin^2(2\theta)}}{2-\sin(2\theta)}. $$

As you see, this is not of the standard form, because non of foci placed in the pole.

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  • $\begingroup$ The discriminant should be $\cos^2\theta+16-8\sin 2\theta$. $\endgroup$ – Blue May 15 at 12:12

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