Question

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

Answer 1

Substitute:

`x=rcosθ`
`y=rsinθ`

Polar equation is

`-rcosθ*rsinθ=-(rcosθ)^2+4(rsinθ)^2`

or, with some trigonometric identities:

`10sin^2θ+sin2θ-2=0`

Explanation:

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Therefore:

`x=rcosθ`

`y=rsinθ`

So, by substituting:

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

`-rcosθ*rsinθ=-(rcosθ)^2+4(rsinθ)^2`

`-r^2cosθsinθ=-r^2cos^2θ+4r^2sin^2θ`

Divide by `r^2` since `r>0`

`-cosθsinθ=-cos^2θ+4sin^2θ`

Using trigonometric identities:

`sin2θ=2sinθcosθ`

`sin^2θ+cos^2θ=1`

We have:

`-(sin2θ)/2=-(1-sin^2θ)+4sin^2θ`

`-(sin2θ)/2=-1+sin^2θ+4sin^2θ`

`-(sin2θ)/2=5sin^2θ-1`

`10sin^2θ+sin2θ-2=0`