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

Jan 20, 2017

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

#### Explanation:

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

then if

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

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

${B}^{2} - 4 A C < 0$ and $A \ne C$, it is an ellipse

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

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

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

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

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