In[28]:= r Sin[\[Theta]] - r^2 Cos[\[Theta]]^2 - 2 == 0

Out[28]= -2 - r^2 Cos[\[Theta]]^2 + r Sin[\[Theta]] == 0

In[30]:= Solve[-2 - r^2 Cos[\[Theta]]^2 + r Sin[\[Theta]] ==  0, {r, \[Theta]}]

Out[30]= {{r ->1/2 (-Sqrt[1 - 9 Cos[\[Theta]]^2] - Sqrt[
      1 - Cos[\[Theta]]^2]) Sec[\[Theta]]^2}, {r -> 
   1/2 (Sqrt[1 - 9 Cos[\[Theta]]^2] - Sqrt[
      1 - Cos[\[Theta]]^2]) Sec[\[Theta]]^2}, {r -> 
   1/2 (-Sqrt[1 - 9 Cos[\[Theta]]^2] + Sqrt[
      1 - Cos[\[Theta]]^2]) Sec[\[Theta]]^2}, {r -> 
   1/2 (Sqrt[1 - 9 Cos[\[Theta]]^2] + Sqrt[
      1 - Cos[\[Theta]]^2]) Sec[\[Theta]]^2}}

In[45]:= PolarPlot[{1/
   2 (-Sqrt[1 - 9 Cos[\[Theta]]^2] - Sqrt[
     1 - Cos[\[Theta]]^2]) Sec[\[Theta]]^2, 
  1/2 (Sqrt[1 - 9 Cos[\[Theta]]^2] - Sqrt[
     1 - Cos[\[Theta]]^2]) Sec[\[Theta]]^2, 
  1/2 (-Sqrt[1 - 9 Cos[\[Theta]]^2] + Sqrt[
     1 - Cos[\[Theta]]^2]) Sec[\[Theta]]^2, 
  1/2 (Sqrt[1 - 9 Cos[\[Theta]]^2] + Sqrt[
     1 - Cos[\[Theta]]^2]) Sec[\[Theta]]^2}, {\[Theta], -3, 3}, 
 AspectRatio -> 1/2]

Take x^2-y+2=a and later make a go to zero

In[13]:= x^2 - y^2 == a /. 
  Thread[Rule[{x, y}, FromPolarCoordinates[{r, \[Theta]}]]];
Solve[%, r]

Out[14]= {{r -> -(Sqrt[a]/Sqrt[
    Cos[\[Theta]]^2 - Sin[\[Theta]]^2])}, {r -> Sqrt[a]/Sqrt[
   Cos[\[Theta]]^2 - Sin[\[Theta]]^2]}}

You can see a approach zero in this manipulate:

Manipulate[
 PolarPlot[a/Sqrt[
  Cos[\[Theta]]^2 - Sin[\[Theta]]^2], {\[Theta], 0, 2 \[Pi]}, 
  PlotPoints -> 1000, PlotRange -> 5], {{a, 0.5}, 0.02, 1, 
  Appearance -> "Labeled"}]

Im a noob to this. I just typed in "x^2-y+2=0 Convert to polar" and its just popping up as a graph (helpful but not exactly what im looking for) Im wondering if it will somehow be able to give me an answer in r= some term of sin(theta)/cos(theta)