How do you write this equation $4/3x^2+4/3y^2=1?$ in standard form and find the center and radius?

Question

2244 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-04T02:34:02

You can rewrite this as

$4/3x^2+4/3y^2=1=>4*(x^2+y^2)=3=>x^2+y^2=(sqrt3/2)^2$

Hence it is a circle with center at $(0,0)$ and radius $r=sqrt3/2$

How do you write this equation 4/3x^2+4/3y^2=1?4/3x^2+4/3y^2=1? in standard form and find the center and radius?