How do you classify the conic #x^2+4y^2-8x+3y+12=0#?

1 Answer
Jan 20, 2017

Answer:

#x^2+4y^2-8x+3y+12=0# is an ellipse.

Explanation:

Let the equation be of the type #Ax^2+Bxy+Cy^2+Dx+Ey+F=0#

then if

#B^2-4AC=0# and #A=0# or #C=0#, it is a parabola

#B^2-4AC<0# and #A=C#, it is a circle

#B^2-4AC<0# and #A!=C#, it is an ellipse

#B^2-4AC>0#, it is a hyperbola

In the given equation #x^2+4y^2-8x+3y+12=0#

#A=1#, #B=0# and #C=4#

Therefore, #B^2-4AC=0^2-4xx1xx4=-16<0# and #A!=C#

Hence, #x^2+4y^2-8x+3y+12=0# is an ellipse.
graph{x^2+4y^2-8x+3y+12=0 [1.302, 6.302, -1.64, 0.86]}