Question

How do you convert r=2sec^2(theta/2) into cartesian form?

Answer 1

The Cartesian equation is `y^2=-8(x-2)`.

Explanation:

To convert the equation, use the reciprocal definition:

`sectheta=1/costheta`

and the cosine half-angle formula:

`cos(theta/2)=+-sqrt((1+costheta)/2)`

Here's the equation:

`r=2sec^2(theta/2)`

`r=2*1/cos^2(theta/2)`

`r=2*1/(cos(theta/2))^2`

`r=2*1/(+-sqrt((1+costheta)/2))^2`

`r=2*1/((1+costheta)/2)`

`r=4/(1+costheta)`

`r+rcostheta=4`

Using the substitutions `r=sqrt(x^2+y^2)` and `rcostheta=x`:

`sqrt(x^2+y^2)+x=4`

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

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

`color(red)cancelcolor(black)(x^2)+y^2=16-8x+color(red)cancelcolor(black)(x^2)`

`y^2=-8x+16`

`y^2=-8(x-2)`

That's the equation; it's leftward-opening parabola. Here's the graph:

graph{y^2=-8(x-2) [-25.66, 25.65, -12.83, 12.83]}

Hope this helped!