# How do you classify the conic -2y^2+x+5y+26=0?

Jul 16, 2016

A parabola

#### Explanation:

Conics can be represented as

$p \cdot M \cdot p + \left\langlep , \left\{a , b\right\}\right\rangle + c = 0$

where $p = \left\{x , y\right\}$ and

$M = \left(\begin{matrix}{m}_{11} & {m}_{12} \\ {m}_{21} & {m}_{22}\end{matrix}\right)$.

For conics ${m}_{12} = {m}_{21}$ then $M$ eigenvalues are always real because the matrix is symetric.

The characteristic polynomial is

$p \left(\lambda\right) = {\lambda}^{2} - \left({m}_{11} + {m}_{22}\right) \lambda + \det \left(M\right)$

Depending on their roots, the conic can be classified as

1) Equal --- circle
2) Same sign and different absolute values --- ellipse
3) Signs different --- hyperbole
4) One null root --- parabola

In the present case we have

$M = \left(\begin{matrix}0 & 0 \\ 0 & - 2\end{matrix}\right)$

with characteristic polynomial

${\lambda}^{2} + 2 \lambda = 0$

with roots $\left\{0 , - 2\right\}$ so we have a parabola.

Determining $\left\{a , b , c\right\}$ is straightforward, solving the conditions for equating coefficients in

$- 2 {y}^{2} + x + 5 y + 26 - \left(p \cdot M \cdot p + \left\langlep , \left\{a , b\right\}\right\rangle + c\right) = 0 \forall x \in \mathbb{R}$

giving

$\left\{a = 1 , b = 5 , c = 26\right\}$