SOLUTION: write in standard form. state wheather the graph is a parabola, circle,ellipse, or hyperbola. and graph the equation...? 6x^2+ 6y^2=162

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: write in standard form. state wheather the graph is a parabola, circle,ellipse, or hyperbola. and graph the equation...? 6x^2+ 6y^2=162      Log On


   



Question 93858: write in standard form. state wheather the graph is a parabola, circle,ellipse, or hyperbola. and graph the equation...?

6x^2+ 6y^2=162

Answer by checkley75(3666) About Me  (Show Source):
You can put this solution on YOUR website!
6x^2+6y^2=162 reducing this equation by diviving all terms by 6 we get
x^2+y^2=27 now settong x=0 & solve for y we get
0^2+y^2=27
y^2=27
y=sqrt27
y=+/-5.196 answer for the y intercepts of(0,5.196)(0,-5.196)
Now set y=0 & solve for x we get
x^2+0^2=27
x^2=27
x=sqrt27
x=+/-5.196 answer for the x intercepts of (5.196,0)(-5.196,0)
Thus one could conclude that this is a circle with an origin of (0,0) & a radius of 5.196 units.