Going from what you did:
... (r^2cos^2θ)-(rsinθ)=0
=> rcos^2θ - sinθ = 0 .............divided both sides by r
=> rcos^2θ = sinθ ..................add sinθ to both sides
=> r = sinθ/cos^2θ .................divide both sides by cos^2θ
=> r = sinθsec^2θ ..................change 1/cos^2θ to sec^2θ by trig identity
Hello, shadowsiren!
Convert y = x² to polar form.
We have: . . . . . . x² .= .y
Substitute: .r²·cos²θ .= .r·sinθ
Divide by r: . r·cos²θ .= .sinθ
Divide by cos²θ: . . r .= .sinθ/cos²θ
Therefore: . . . . . . r .= .secθ·tanθ
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
We have: . . . . . . . . . . .x² - y² .= .25
Substitute: .(r·cosθ)² - (r·sinθ)² .= .25
Then: . . . . . r²·cos²θ - r²·sin²θ .= .25
Factor: . . . . . r²(cos²θ - sin²θ) .= .25
and we have: . . . . . . r²·cos2θ .= .25
Therefore: . . . . . . . . . . . . .r² .= .25·sec2θ