Question

Can someone help me understand this equation? (writing a polar equation of a conic)

How does a conic with an eccentricity of 4/5 and a directrix of x=3 and a focus at the pole become `12/(5+costheta)`?
The closest I got was `(3(4/5))/(1+(4/5)costheta` which I simplified to `12/(9costheta)`

Answer 1

`r = 12/{4 cos theta + 5}`

Explanation:

A conic with eccentricity `e=4/5` is an ellipse.

For every point on the curve the distance to the focal point over the distance to the directrix is `e=4/5.`

Focus at the pole? What pole? Let's assume the asker means focus at the origin.

Let's generalize the eccentricity to `e` and the directrix to `x=k`.

The distance of a point `(x,y)` on the ellipse to the focus is

`\sqrt{x^2 + y^2}`

The distance to the directrix `x=k` is `|x-k|`.

`e = \sqrt{x^2 + y^2}/|x-k|`

`e^2 = {x^2 + y^2}/(x-k)^2`

That's our ellipse, there's no particular reason to work it into standard form.

Let's make it polar, `r^2=x^2+y^2` and `x=r cos theta`

`e^2 = r^2 /(r cos theta -k)^2`

`e^2(r cos theta - k)^2 = r^2`

`(e r cos theta - e k)^2 - r^2 = 0`

`(r e cos theta + r - ek)(r e cos theta - r - ek) = 0`

`r = {ek}/{e cos theta + 1 } or r = {ek}/{e cos theta - 1 }`

We drop the second form because we never had negative `r`.

So the polar form for an ellipse with eccentricity `e` and directrix `x=k` is

`r = {ek}/{e cos theta + 1 }`

That seems to be the form you started from.

Plugging in `e=4/5, k=3`

`r = {12/5}/{4/5 cos theta + 1 }`

Simplifying gives,

`r = 12/{4 cos theta + 5}`

That's none of the above.