Question

How do you convert `3=(x+4y)^2+3x` into polar form?

Answer 1

Please see the explanation for a detailed description of how the conversion is done.

Explanation:

Expand the square, using the F.O.I.L. method:

`3 = x^2 + 8xy + 16y^2 + 3x`

Substitute `rcos(theta)` for x and `rsin(theta)` for y:

`3 = (rcos(theta))^2 + 8(rcos(theta))(rsin(theta)) + 16(rsin(theta))^2 + 3(rcos(theta))`

This can be written as a quadratic in r:

`0 = (cos^2(theta) + 8cos(theta)sin(theta) + 16sin^2(theta))r^2 + 3cos(theta)r - 3`

The coefficient for `r^2` factors into a square:

`0 = (cos(theta) + 4sin(theta))^2r^2 + 3cos(theta)r - 3`

Use the quadratic formula:

`r = (-b +-sqrt(b^2 - 4(a)(c)))/(2a)`

where:

`a = (cos(theta) + 4sin(theta))^2`
`b = 3cos(theta)`
`c = -3`

Also, we must change the `+-` to only `+`, because negative values for r do not make sense.

`r = (-3cos(theta) +sqrt(9cos^2(theta) + 12(cos(theta) + 4sin(theta))^2))/(2(cos(theta) + 4sin(theta))^2)`