How do you know if the conic section #x^2 +4y^2-8x -48= 0# is a parabola, an ellipse, a hyperbola, or a circle?

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Answer 1 · 2016-09-13T13:13:21

#x^2+4y^2-8x-48=0# is an ellipse.

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-48=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-48=0# is an ellipse.

graph{x^2+4y^2-8x-48=0 [-6.71, 13.29, -4.52, 5.48]}