How do you know if #x^2+2y^2+2x-12y+11=0# is a hyperbola, parabola, circle or ellipse?

1 Answer
Jul 19, 2018

Answer:

Equation of the ellipse # (x+1)^2/(2 sqrt2)^2+( y-3)^2/2^2=1#

Explanation:

# x^2+2 y^2+2 x -12 y+11=0#

The general equation of conics is

#ax^2+bxy+c y^2+d x +e y +f=0 # , for ellipse or circle

#b^2-4 ac <0# for parabola , #b^2-4 ac = 0# and hyperbola,

#b^2-4 ac > 0 ; a= 1, b= 0 , c=2 :. b^2-4a c <0 :.#circle or

ellipse. # (x+1)^2+2( y-3)^2=19-11# or

# (x+1)^2+2( y-3)^2=8# or

# (x+1)^2/(2 sqrt2)^2+( y-3)^2/2^2=1#

This is standard equation of horizontal ellipse is

# x^2/a^2+y^2/b^2=1 ; a>0 , b>0 #

Hence this is ellipse.

graph{x^2+2y^2+2x-12y+11=0 [-20, 20, -10, 10]} [Ans]