Question 1204941


{{{r = 4sin(theta)}}}

we know that:

{{{x=r*cos(theta) }}} => {{{cos(theta)=x/r}}}.....eq.1

{{{y=r*sin(theta) }}}=>{{{sin(theta)=y/r}}}....eq.2

{{{tan^-1(theta)=y/x}}}....eq.3

{{{r=sqrt(x^2+y^2)}}} => {{{r^2=x^2+y^2}}}....eq.4



then, substituting {{{sin(theta)}}} from eq.2 given equation  becomes

 {{{r = 4(y/r)}}}......both sides multiply by {{{r}}}

{{{r^2 = 4y}}} ....substitute {{{r^2}}} from eq.4

{{{x^2+y^2 = 4y }}}

{{{x^2+y^2 -4y =0}}} => a Cartesian equation