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

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Answer 1 · 2017-01-20T08:45:44

$x^2+4y^2-8x+3y+12=0$ 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 $x^2+4y^2-8x+3y+12=0$

$A=1$ , $B=0$ 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]}

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