Which conic section has the polar equation #r=a sintheta#?

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Answer 1 · 2014-12-03T03:42:36

Remember:

#{(x=r cos theta),(y=r sin theta):}#

and

#x^2+y^2=r^2#.

By multiplying by #r#,

#r=a sin theta => r^2=a r sin theta#

by rewriting in rectangular coordinates,

#=>x^2+y^2=ay => x^2+y^2-ay=0#

by adding #(a/2)^2#,

#=> x^2+y^2-ay+(a/2)^2=(a/2)^2#

#x^2+(y-a/2)^2=(a/2)^2#

Hence, it is a circle with radius #a/2#, centered at #(0,a/2)#.

I hope that this was helpful.