How do you identity if the equation x^2-8y+y^2+11=0x28y+y2+11=0 is a parabola, circle, ellipse, or hyperbola and how do you graph it?

2 Answers
Nov 9, 2016

Circle

Explanation:

Modify the equation by completing the square for yy.

x^2 + (y-4)^2 = 5x2+(y4)2=5

The equation above describes a circle of radius sqrt55 centered at (0,4)(0,4).

graph{x^2 + (y-4)^2 = 5 [-9.67, 10.33, -1.6, 8.4]}

Nov 9, 2016

the eqn of a circle is (x-a)^2+(y-b)^2+r^2(xa)2+(yb)2+r2

when multiplied and simplified out the eqn can be rearranged to the form

x^2+y^2+fx+gy+h=0x2+y2+fx+gy+h=0

note;

1) the coefficients of xx &yy are the same.

2) there are no 'xy' terms

Ellipses are of the form

x^2/a+y^2/b=1

hyperbolas are of the form

x^2/a-y^2/b=1

again no 'xy' terms.