# How do you find the standard form of 25x^2+4y^2-250x+16y+640=0 and what kind of a conic is it?

Jul 8, 2016

A parabola

$\left\{x - 5 , y + 2\right\} \cdot \left(\begin{matrix}25 & 0 \\ 0 & 4\end{matrix}\right) \cdot \left\{x - 5 , y + 2\right\} - 1$

#### Explanation:

Given a conic $f \left(x , y\right)$ we can classify it studing the eigenvalues of its associated quadratic form. This quadratic form can be calculated as

${q}_{f} \left(x , y\right) = \frac{1}{2} \left\{x , y\right\} \cdot \left(\begin{matrix}{f}_{x x} & {f}_{x y} \\ {f}_{y x} & {f}_{y y}\end{matrix}\right) \cdot \left\{x , y\right\}$

Here ${f}_{{x}_{1} , {x}_{2}} = \frac{{\partial}^{2} f}{\partial {x}_{1} \partial {x}_{2}}$

In this case we have

${q}_{f} \left(x , y\right) = \left\{x , y\right\} \cdot \left(\begin{matrix}25 & 0 \\ 0 & 4\end{matrix}\right) \cdot \left\{x , y\right\}$

with eigenvalues $\left\{25 , 4\right\}$ characterizing it as a parabola.

The standard form is

${f}_{s} \left(x , y\right) = \left\{x - {x}_{0} , y - {y}_{0}\right\} \cdot \left(\begin{matrix}25 & 0 \\ 0 & 4\end{matrix}\right) \cdot \left\{x - {x}_{0} , y - {y}_{0}\right\} + {c}_{0}$

Doing $f \left(x , y\right) - {f}_{s} \left(x , y\right) = 0$ and equating the coefficients we have

{(640 - c_0 - 25 x_0^2 - 4 y_0^2 = 0), (16 + 8 y_0 = 0), (50 (x_0-5) =0) :}

solving for ${x}_{0} , {y}_{0} , {c}_{0}$ we get

${f}_{s} \left(x , y\right) = \left\{x - 5 , y + 2\right\} \cdot \left(\begin{matrix}25 & 0 \\ 0 & 4\end{matrix}\right) \cdot \left\{x - 5 , y + 2\right\} - 1$