Question

How do you write the Cartesian Equation `r^2 = sin 2(theta)`?

Answer 1

`(x^2+y^2)^2-2xy=0`

Explanation:

Use `r^2=x^2+y^2, sin theta = y/r and cos theta = x/r`.

Here, `x^2+y^2=2sin theta cos theta=2(xy)/r^2=(2xy)/(x^2+y^2)`.

So, `(x^2+y^2)^2-2xy=0`.

The curve is looking like `oo` rotated through

through `45^0`, in the anticlockwise sense, about the center (origin)..

The graph of the Limacon `r^2=cos 2theta` looks like `oo`.