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White raven [17]
3 years ago
13

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

Mathematics
1 answer:
11Alexandr11 [23.1K]3 years ago
5 0
\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
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