How do you convert $5 y = {(x + y)}^{2} - 2 x y$ $5y=(x+y)^2 -2xy$ into a polar equation?

Question

1566 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-15T05:41:39

$r = 5 sin θ$ $r=5sintheta$

The relationn between Cartesian or rectangular coordinates $(x, y)$ $(x,y)$ and polar coordinates $(r, θ)$ $(r,theta)$ is given by

$x = r cos θ$ $x=rcostheta$ , $y = r sin θ$ $y=rsintheta$ and hence $x^{2} + y^{2} = r^{2}$ $x^2+y^2=r^2$

Therefore we can write $5 y = {(x + y)}^{2} - 2 x y$ $5y=(x+y)^2-2xy$ as

$5 r sin θ = x^{2} + y^{2} + 2 x y - 2 x y$ $5rsintheta=x^2+y^2+2xy-2xy$

or $5 r sin θ = x^{2} + y^{2}$ $5rsintheta=x^2+y^2$

or $5 r sin θ = r^{2}$ $5rsintheta=r^2$

or $r = 5 sin θ$ $r=5sintheta$

This is the equation of a circle

graph{5y=(x+y)^2-2xy [-10.13, 9.87, -2.92, 7.08]}

How do you convert 5y=(x+y)^2 -2xy 5y=(x+y)2−2xy5y=(x+y)^2 -2xy into a polar equation?