# polar form of ellipse

• May 3rd 2009, 06:41 AM
Stonehambey
polar form of ellipse
The problem is to prove that the following curve is an ellipse

$r=\frac{2a}{3+2\cos\theta}$

I've looked about and have found the equation of the ellipse in polar form, but I can't find it's derivation or proof anywhere.

Anyway, I started this problem by subbing in

$r = \sqrt{x^2+y^2}$

and

$x = r\cos\theta$

But I ended up with a horrible mess and couldn't spot how to get it into the form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

Any help or guidance would be much appreciated :)

Stonehambey
• May 3rd 2009, 07:47 AM
Soroban
Hello, Stonehambey!

Quote:

Prove that the following curve is an ellipse:

. . $r\:=\:\frac{2k}{3+2\cos\theta}$

You have the right idea, but I've learned that it's better to simplify first.

We have: . $r(3 + 2\cos\theta) \:=\:2k \quad\Rightarrow\quad 3r + 2r\cos\theta \:=\:2k \quad\Rightarrow\quad 3r \:=\:2k - 2r\cos\theta$

Substitute now: . $3\underbrace{r} \;=\;2k - 2\underbrace{r\cos\theta}$
. . . . . . . . . . . . . . $\downarrow\qquad\qquad\qquad\;\; \downarrow$
. . . . . . . . . . $3\sqrt{x^2+y^2} \;\;=\;\; 2k - 2x$

Square both sides: . $9(x^2+y^2) \;=\;(2k-2x)^2$

. . . . . . $9x^2 + 9y^2 \;=\;4k^2 - 8kx + 4x^2 \quad\Rightarrow\quad 5x^2 + 8kx + 9y^2 \;=\;4k^2$

Complete the square: . $5\left(x^2 + \tfrac{8}{5}x\right) + 9y^2 \;=\;4k^2$

. . $5\left(x^2 + \tfrac{8}{5}kx + {\color{red}\tfrac{16}{25}k^2}\right) + 9y^2 \;=\;4k^2 + {\color{red}\tfrac{16}{5}k^2} \quad\Rightarrow\quad 5\left(x + \tfrac{4}{5}k\right)^2 + 9y^2 \;=\;\tfrac{36}{5}k^2$

Divide by $\tfrac{36}{5}k^2\!:\quad\frac{5(x+\frac{4}{5}k)^2} {\frac{36}{5}k^2} + \frac{9y^2}{\frac{36}{5}k^2} \;=\;1$

And we have: . $\frac{(x+\frac{4}{5}k)^2}{\frac{36}{25}k^2} + \frac{y^2}{\frac{4}{5}k^2} \;=\;1$

This an ellipse with: . center $\left(\text{-}\tfrac{4}{5}k,\:0\right),\quad a \:=\: \tfrac{6}{5}k,\quad b \:=\:\tfrac{2}{\sqrt{5}}k$

• May 3rd 2009, 08:15 AM
Opalg
Quote:

Originally Posted by Stonehambey
The problem is to prove that the following curve is an ellipse

$r=\frac{2a}{3+2\cos\theta}$

I've looked about and have found the equation of the ellipse in polar form, but I can't find it's derivation or proof anywhere.

Anyway, I started this problem by subbing in

$r = \sqrt{x^2+y^2}$

and

$x = r\cos\theta$

But I ended up with a horrible mess and couldn't spot how to get it into the form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

There are two ways of doing this: geometric or analytic.

Geometrically, an ellipse is defined as the locus of a point which moves so that its distance from a fixed line (the directrix) is a constant multiple (the eccentricity, which must be less than 1) of its distance from a fixed point (the focus). In this case, you can write the equation as $r = \tfrac23(a-x)$. This says that r (the distance of the point [r,θ] from the origin) is 2/3rds of its distance from the line x=a. So the locus is an ellipse, with focus at the origin, directrix the line x=a, and eccentricity 2/3.

To solve the problem analytically, start from the equation in the form $3r = 2(a-x)$. Square both sides: $9(x^2+y^2) = 4(a-x)^2 = 4a^2-8ax+4x^2$. Rearrange this as $5x^5+8ax+9y^2 = 4a^2$. Complete the square: $5(x+\tfrac45a)^2 + 9y^2 = \tfrac{36}{25}a^2 = (\tfrac65a)^2$. Finally, divide both sides by $(\tfrac65a)^2$ and you get the equation in the form $\frac{(x+\frac45a)^2}{(\frac65a)^2} + \frac{y^2}{(\frac2{\sqrt5}a)^2} = 1$. Again, this is the equation of an ellipse. It is not centred at the origin, but at the point $(-\tfrac45a,0)$. Its semi-axes are $\tfrac65a$ and $\tfrac2{\sqrt5}a$.

Edit. Soroban got there first with the analytic version!