# generic parabola in polar coordinates

Starting from the equation $$y=ax^2+bx+c$$ substituting I get the next equation in polar coordinates: $$a\cos^2 \theta\ \rho^ 2 + (b \cos \theta - \sin \theta)\ \rho + c = 0$$ in case C was $$0$$ we could explicate ρ, so I made sure that c would result $$0$$, by find an auxiliary Cartesian plane of cordinate $$(u, v)$$ so that the equation of the parabola became $$v=au^2+bu$$ Now I can find the polar coordinates with respect to the new pole, $$( u=\rho_1\cos\theta_1 , v=\rho_1\sin\theta_1 )$$ So the equation of the parabola in this case becomes: $$\rho_1\sin(\theta_1)=a(\rho_1\cos(\theta_1))^2+b\rho_1\cos(\theta_1)$$ explaining $$\rho_1$$ we get: $$\rho_1=\frac{\sec\theta_1(\tan\theta_1-b)}{a}$$ $$a\ne0$$ $$\theta\ne0$$

To return to ρ we can use the law of cosines $$\rho=\sqrt{c^2+\rho_1^2-2c\rho_1\cos\alpha}$$ but $$\alpha=90-\theta$$ so $$\rho=\sqrt{c^2+\rho_1^2-2c\rho_1\sin\theta}$$ definitely $$\rho=\sqrt{c^2+(\frac{\sec\theta(\tan\theta-b)}{a})^2-\frac{2c\tan\theta(\tan\theta-b)}{a}}$$ Do you think it's correct?

• Can anyone help me? – Leprep98 Feb 9 at 11:19
• I haven't checked, but don't understand why you don't solve for $\rho$ the original quadratic equation. – Aretino Feb 9 at 13:04
• @Aretino I have had this idea so I have wanted to try this way – Leprep98 Feb 9 at 14:11
• The problem is you wrote $\theta$ for $\theta_1$ in your last expression, which is not correct because those angles are not equal. And converting $\theta_1$ to $\theta$ is not easy. – Aretino Feb 9 at 14:43
• @Aretino thanks you, now I see the mistake. – Leprep98 Feb 9 at 15:06