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

1 Answer
Jul 8, 2016

A parabola

{x-5,y+2}cdot((25,0),(0,4))cdot {x-5,y+2}-1

Explanation:

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

q_f(x,y) =1/2 {x,y}cdot((f_{x x},f_{xy}),(f_{yx},f_{yy}))cdot {x,y}

Here f_{x_1,x_2} = (partial^2f)/(partial x_1partial x_2)

In this case we have

q_f(x,y) = {x,y}cdot((25,0),(0,4))cdot {x,y}

with eigenvalues {25,4} characterizing it as a parabola.

The standard form is

f_s(x,y) = {x-x_0,y-y_0}cdot((25,0),(0,4))cdot {x-x_0,y-y_0}+c_0

Doing f(x,y)-f_s(x,y)=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(x,y) = {x-5,y+2}cdot((25,0),(0,4))cdot {x-5,y+2}-1