Question

What is the Cartesian form of `2r = -tan^2theta+4sec^2theta`?

Answer 1

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

Explanation:

From the pass equations
`x = r cos(theta)`
`y = r sin(theta)`
and making the substitutions
`r = x/cos(theta), theta = arctan(y/x)`
we obtain the Cartesian form.