Question

Convert: 6y + y² = x² from rectangular to polar form.

Answer 1

$\bf \textit{Double Angle Identities} \\ \quad \\ sin(2\theta)=2sin(\theta)cos(\theta) \\ \quad \\\\ cos(2\theta)= \begin{cases} \boxed{cos^2(\theta)-sin^2(\theta)}\\ 1-2sin^2(\theta)\\ 2cos^2(\theta)-1 \end{cases}\\\\\\ tan(2\theta)=\cfrac{2tan(\theta)}{1-tan^2(\theta)}\\\\ -------------------------------$

$\bf 6y+y^2=x^2\implies \cfrac{6y}{x^2}+\cfrac{y^2}{x^2}=1\implies \cfrac{6rsin(\theta )}{[rcos(\theta )]^2}+\cfrac{[rsin(\theta )]^2}{[rcos(\theta )]^2}=1 \\\\\\ \cfrac{6rsin(\theta )}{r^2cos^2(\theta )}+\cfrac{r^2sin^2(\theta )}{r^2cos^2(\theta )}=1\implies \cfrac{6sin(\theta )}{rcos^2(\theta )}+\cfrac{sin^2(\theta )}{cos^2(\theta )}=1$

$\bf \cfrac{6sin(\theta )}{rcos^2(\theta )}=1-\cfrac{sin^2(\theta )}{cos^2(\theta )}\implies \cfrac{6sin(\theta )}{1-\frac{sin^2(\theta )}{cos^2(\theta )}}=rcos^2(\theta ) \\\\\\ \cfrac{6sin(\theta )}{cos^2(\theta )\left[ 1-\frac{sin^2(\theta )}{cos^2(\theta )} \right]}=r\implies \cfrac{6sin(\theta )}{cos^2(\theta )-sin^2(\theta )}=r \\\\\\ \cfrac{6sin(\theta )}{cos(2\theta )}=r$