Question

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

Answer 1

`4(x^2+y^2)^2=x^2(x^2+y^2)sqrt(x^2+y^2).` A graph is inserted.

Explanation:

The conversion formula is `r(cos theta, sin theta)=(x, y)`, giving

`cos theta = x/r and sin theta = y/r`, where `r = sqrt(x^2+y^2)`.

Substitution and reorganization gives the answer.

As r is a function of both the squares of sine and cosine, the graph

is symmetrical about `theta = 0 and theta = pi/2` ( x-axis and y-axis).

graph{(x^2+y^2)^1.5(4sqrt(x^2+y^2)-x^2)-x^2y^2=0 [-80, 80, -40, 40]}

Answer 2

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

Explanation:

The relation between polar coordinates `(r.theta)` and Cartesian coordinates `(x,y)` is given by

`x=rcostheta`, `y=rsintheta`, `r^2=x^2+y^2` and `tantheta=y/x`

Hence `r=-sin^2theta+4sec^2theta`

or `r=-y^2/r^2+(4xxr^2/x^2)`

or `r^3x^2=-x^2y^2+4r^4`

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