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

Answer:

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#