# How do you identify the type of conic 4x^2+8y^2-8x-24=4 is, if any and if the equation does represent a conic, state its vertex or center?

Jul 16, 2016

An ellipse

#### 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 --- hyperbola
4) One null root --- parabola

In the present case we have

$M = \left(\begin{matrix}4 & 0 \\ 0 & 8\end{matrix}\right)$

with characteristic polynomial

${\lambda}^{2} - 12 \lambda + 32 = 0$

with roots $\left\{4 , 8\right\}$ so we have an ellipse.

Being an ellipse there is a canonical representation for it

${\left(\frac{x - {x}_{0}}{a}\right)}^{2} + {\left(\frac{y - {y}_{0}}{b}\right)}^{2} = 1$

${x}_{0} , {y}_{0} , a , b$ can be determined as follows

$4 {x}^{2} + 8 {y}^{2} - 8 x - 28 - \left({b}^{2} {\left(x - {x}_{0}\right)}^{2} + {a}^{2} {\left(y - {y}_{0}\right)}^{2} - {a}^{2} {b}^{2}\right) = 0 \forall x \in \mathbb{R}$

giving

{ (-28 + a^2 b^2 - b^2 x_0^2 - a^2 y_0^2 = 0), (2 a^2 y_0 = 0), (8 - a^2 = 0), (-8 + 2 b^2 x_0 = 0), (4 - b^2 = 0) :}

solving we get

$\left\{{a}^{2} = 8 , {b}^{2} = 4 , {x}_{0} = 1 , {y}_{0} = 0\right\}$

so

$\left\{4 {x}^{2} + 8 {y}^{2} - 8 x - 24 = 4\right\} \equiv \left\{{\left(x - 1\right)}^{2} / 8 + {y}^{2} / 4 = 1\right\}$ 