How do you identity if the equation #3x^2+4y^2+8y=8# is a parabola, circle, ellipse, or hyperbola and how do you graph it?

1 Answer
Nov 19, 2016

Answer:

This is an ellipse

Explanation:

Compare this to the general equation for conics and calculate the discriminant.

#Ax^2+Bxy+Cy^2+Dx+Ey+F=0#

#3x^2+4y^2+8y-8=0#

The discriminant is

#Delta=B^2-4AC#

if #Delta<0#, we have an ellipse

If #Delta=0#, we have a parabola

If #Delta>0#, we have a hyperbola

In our case,

#Delta=0-4*3*4=-48#

so, #Delta<0#, we have an ellipse.

We can also, rearrage the equation

#3x^2+4(y^2+2y)=8#

#3x^2+4(y^2+2y+1)=8+4#

#3x^2+4(y+1)^2=12#

Dividing by 12

#x^2/4+(y+1)^2/3=1#

which is an equation of an ellipse

The center is #(0,-1)#

#c=sqrt(4-3)=1#

The foci are F #=(1,-1)# and F'#=(-1,-1)#

The major axis is #=4#

and the minor axis is #=sqrt3#

graph{(3x^2+4y^2+8y-8)(y-1)=0 [-4.915, 3.855, -3.362, 1.023]}