Question

How do you convert `r = 6/(3 - 4cos(theta))` into cartesian form?

Answer 1

`7x^2-9y^2+48x+36=0`

Explanation:

Use `(x, y) = r (cos theta, sin theta) and r=sqrt(x^2+y^2)`,

Here, cross-multiplying and rearranging,

`3r=3sqrt(x^2+y^2)=6+4r cos theta=6+4x`. Squaring,

`9(x^2+y^2)=(6+4x)^2=36+48x+16x^2`. Rearranging,

`7x^2-9y^2+48x+36=0`

This represents a hyperbola with eccentricity`e = 4/3`.

Answer 2

`7x^2-9y^2+48x+36=0`

Explanation:

We know that the cartesian coordinate `(x,y)` of a point is related with its polar coordinate `(r,theta)` as follows:

`x=rcostheta and y=rsintheta->r=sqrt(x^2+y^2)`

The given equation

`r=6/(3-4costheta)`

`=>3r-4rcostheta=6`

`=>3sqrt(x^2+y^2)-4x=6`

`=>(3sqrt(x^2+y^2))^2=(6+4x)^2`

`=>9x^2+9y^2=36+48x+16x^2`

`=>16x^2-9x^2+48x-9y^2+36=0`

`=>7x^2-9y^2+48x+36=0`

This is the cartesian form of the given polar equation.

graph{7x^2-9y^2+48x+36=0}

The source for what follows is A. S. Adikesavan.

For information, the polar equation

`r = d/(sqrt(a^2 + b^2) + c(a cos theta + b sin theta))`

represents

( parabola ellipse hyperbola) according as

( abs c =1 abs c < 1 abs c > 1).

For d = 2, a = 2, b = 2, c = 2 giving

`r =`1/(1 + 2(cos theta + sin theta))` that represents a hyperbola.

graph{ ((x^2+y^2)^0.5-1 +2(x + y))( -(x^2+y^2)^0.5-1 +2(x + y)) = 0[-2 2 -2 2]}

Changing c to 1, in the above assignment, a parabola is traced.

graph{ (x^2+y^2)^0.5-1 +(x + y) = 0[-2 2 -2 2]}

Changing c to 0.5, in the above assignment, an ellipse is traced.

graph{ (x^2+y^2)^0.5-1 +0.5(x + y) = 0[-4 2 -4 2]}