Question

convert the equation into polar form
y=8

Answer 1

\[put x=r \cos \theta =0 \]
\[ Put y=r \sin \theta =8 \]
square and add
\[x ^{2}+y ^{2}=r ^{2}\left( \cos ^{2} \theta+\sin ^{2}\theta\right)=8^{2},r=8\]
\[divide,\frac{ y }{ x }=\frac{ r \sin \theta }{r \cos \theta }=\frac{ 8 }{0 }=\infty\]
\[\tan \theta=\infty=\tan \frac{ \pi }{ 2 },\theta=\frac{ \pi }{ 2 }\]
\[hence polar co ordinates are \left( r,\theta \right) i.e.,\left( 8,\frac{ \pi }{ 2 } \right)\]

Answer 2

\[y=r\sin\theta~\Rightarrow~8=r\sin\theta\]
Now write that as a function of \(r\) in terms of \(\theta\).

Answer 3

thanks guys!!