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

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Answer 1 · 2017-01-20T08:45:44

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

Let the equation be of the type $A x^{2} + B x y + C y^{2} + D x + E y + F = 0$ $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$

then if

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

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

$B^{2} - 4 A C < 0$ $B^2-4AC<0$ and $A \neq C$ $A!=C$ , it is an ellipse

$B^{2} - 4 A C > 0$ $B^2-4AC>0$ , it is a hyperbola

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

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

Therefore, $B^{2} - 4 A C = 0^{2} - 4 \times 1 \times 4 = - 16 < 0$ $B^2-4AC=0^2-4xx1xx4=-16<0$ and $A \neq C$ $A!=C$

Hence, $x^{2} + 4 y^{2} - 8 x + 3 y + 12 = 0$ $x^2+4y^2-8x+3y+12=0$ 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=0x2+4y2−8x+3y+12=0x^2+4y^2-8x+3y+12=0?