Question

How do you convert `y= -2x^2+3xy-4` into a polar equation?

Answer 1

We know the relations

`x=rcostheta and y =rsintheta`

Again `x^2+y^2=r^2`

where r and `theta` are the polar coordinate of a point having rectangular coordinate `(x,y)`

The given equation in rectanglar form is

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

`=>rsintheta=-2r^2cos^2theta+3r^2sinthetacostheta-4`

`=>rsintheta+2r^2cos^2theta-3r^2sinthetacostheta+4=0`

This is the polar form of the given equation.

Answer 2

Substitute `rcos(theta)` for `x` and `rsin(theta)` for `y`:
`r = (-sin(theta) +-sqrt(sin^2(theta) + 16(3cos(theta)sin(theta) - 2cos^2(theta))))/((6cos(theta)sin(theta) - 4cos^2(theta)))`

Explanation:

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

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

`r^2(3cos(theta)sin(theta) - 2cos^2(theta)) - rsin(theta) - 4`

This a quadratic of the form `ar^2 + br + c` where `a = (3cos(theta)sin(theta) - 2cos^2(theta))`, `b = -sin(theta)`, and `c = -4`

Using the quadratic formula, `r = (-b +-sqrt(b^2 - 4(a)(c)))/(2a)`:

`r = (-sin(theta) +-sqrt(sin^2(theta) + 16(3cos(theta)sin(theta) - 2cos^2(theta))))/((6cos(theta)sin(theta) - 4cos^2(theta)))`