# How do you use the discriminant to classify the conic section x^2 + y^2 - 6x + 8y - 24 = 0?

Ellipse

#### Explanation:

${x}^{2} + {y}^{2} - 6 x + 8 y - 24 = 0$

Comparing above equation with the standard form of quadratic equation: $a {x}^{2} + 2 h x y + b {y}^{2} + 2 g x + 2 f y + c = 0$ we get

$a = 1 , h = 0 , b = 1 , g = - 3 , f = 4 , c = - 24$

Now, using determinant $\left(\setminus \Delta\right)$ of quadratic equation as follows

$\setminus \Delta = a b c + 2 f g h - a {f}^{2} - b {g}^{2} - c {h}^{2}$

$= \left(1\right) \left(1\right) \left(- 24\right) + 2 \left(4\right) \left(- 3\right) \left(0\right) - \left(1\right) {\left(4\right)}^{2} - \left(1\right) {\left(- 3\right)}^{2} - \left(- 24\right) {\left(0\right)}^{2}$

$= - 49$

$\setminus \because \Delta \setminus \ne 0$ hence the given quadratic equation shows a conic section. ($\Delta = 0$ is the case of pair of lines)

Now, using determinant of conic section :

${h}^{2} - a b$

$= {0}^{2} - 1 \setminus \cdot 1$

$= - 1$

$\setminus \because {h}^{2} - a b < 0$ hence the given quadratic equation shows an ellipse .