# Is the following equation an ellipse, circle, parabola, or hyperbola 4x^2-3y^2-48x-6y+129=0?

Mar 30, 2018

The reference Conic Section tells us that given equation:

$4 {x}^{2} - 3 {y}^{2} - 48 x - 6 y + 129 = 0$

Is in the General Cartesian form:

$A {x}^{2} + B x y + C {y}^{2} + D x + E y + F = 0$

where $A = 3 , B = 0 , \mathmr{and} C = - 3$

The same reference, also. tells us that the determinant, ${B}^{2} - 4 A C$ and be used to determine the type of conic section:

Compute the determinant:

${B}^{2} - 4 A C = {0}^{2} - 4 \left(3\right) \left(- 3\right) = 36$

The determinant is greater than zero, therefore, the conic section is a hyperbola. Here is its graph as proof.

graph{4x^2-3y^2-48x-6y+129=0 [-7.37, 21.11, -7.57, 6.67]}