Question 1204937



{{{y^2 = 9x}}}

In polar coordinates, {{{x=r*cos(theta) }}}and {{{y=r*sin(theta)}}}


give equation can be rewritten as 


{{{(r*sin(theta))^2=9r*cos(theta)}}}


{{{r^2*sin^2(theta)=9r*cos(theta)}}}...simplify, divide by {{{r}}}

{{{r*sin^2(theta)=9cos(theta)}}}

{{{r=(9cos(theta))/sin^2(theta)}}}


{{{y^2=9x}}}  in polar coordinates is {{{r=(9cos(theta))/sin^2(theta)}}}