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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?

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  • $\begingroup$ Can anyone help me? $\endgroup$ – Leprep98 Feb 9 at 11:19
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    $\begingroup$ I haven't checked, but don't understand why you don't solve for $\rho$ the original quadratic equation. $\endgroup$ – Aretino Feb 9 at 13:04
  • $\begingroup$ @Aretino I have had this idea so I have wanted to try this way $\endgroup$ – Leprep98 Feb 9 at 14:11
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    $\begingroup$ 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. $\endgroup$ – Aretino Feb 9 at 14:43
  • $\begingroup$ @Aretino thanks you, now I see the mistake. $\endgroup$ – Leprep98 Feb 9 at 15:06

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