How do I graph the ellipse with the equation #x^2+4y^2-4x+8y-60=0#?

1 Answer
Jun 6, 2016

Answer:

An ellipse centered at #(2, –1)# with horizontal radius #sqrt(68)# units and vertical radius #sqrt(17)# units.

Explanation:

Firstly, complete the square.

#x^2 - 4x + 4y^2 + 8y = 60#
#x^2 - 2(2)x + 2^2 + 4(y^2 + 2(1)y + 1^2) = 60 + 2^2 + 4#
#(x-2)^2 + 4(y+1)^2 = 68#
#(x-2)^2/68 + (y+1)^2/17 = 1#
#(x-2)^2/(sqrt(68))^2 + (y+1)^2/(sqrt(17))^2 = 1#

Thus the graph is an ellipse centered at #(2, –1)# with horizontal radius #sqrt(68)# units and vertical radius #sqrt(17)# units.