Question

How do you convert `r(2 - cosx) = 2` to rectangular form?

Answer 1

The explanation is written with the assumption that you meant `theta` for the argument of the cosine function.

Explanation:

Use the distributive property :

`2r - rcos(theta) = 2`

`2r = rcos(theta) + 2`

Substitute x for `rcos(theta)` and `sqrt(x^2 + y^2)` for r:

`2sqrt(x^2 + y^2) = x + 2`

Square both sides:

`4(x^2 + y^2) = x^2 + 4x + 4`

`4x^2 + 4y^2 = x^2 + 4x + 4`

`3x^2 - 4x + 4y^2 = 4`

Add `3h^2` to both sides:

`3x^2 - 4x + 3h^2 + 4y^2 = 3h^2+ 4`

Remove a factor of the 3 from the first 3 terms:

`3(x^2 - 4/3x + h^2) + 4y^2 = 3h^2+ 4`

Use the middle in the right side of the pattern `(x - h)^2 = x^2 - 2hx + h^2` and middle term of the equation to find the value of h:

`-2hx = -4/3x`

`h = 2/3`

Substitute the left side of the pattern into the equation:

`3(x - h)^2 + 4y^2 = 3h^2+ 4`

Substitute `2/3` for h and insert a -0 into the y term:

`3(x- 2/3)^2 + 4(y-0)^2 = 3(2/3)^2+ 4`

`3(x- 2/3)^2 + 4(y-0)^2 = 16/3`

Divide both sides by `16/3`

`3(x- 2/3)^2/(16/3) + 4(y-0)^2/(16/3) = 1`

`(x- 2/3)^2/(16/9) + (y-0)^2/(16/12) = 1`

Write the denominators as squares:

`(x- 2/3)^2/(4/3)^2 + (y-0)^2/(4/sqrt(12))^2 = 1`

The is standard Cartesian form of the equation of an ellipse with a center at `(2/3,0)`; its semi-major axis is `4/3` units long and is parallel to the x axis and its semi-minor axis is `4/sqrt(12)` units long and is parallel to the y axis.