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?

1 Answer
Jul 16, 2016

Answer:

An ellipse

Explanation:

Conics can be represented as

#p cdot M cdot p+<< p, {a,b}>>+c=0#

where #p = {x,y}# and

#M = ((m_{11},m_{12}),(m_{21}, m_{22}))#.

For conics #m_{12} = m_{21}# then #M# eigenvalues are always real because the matrix is symetric.

The characteristic polynomial is

#p(lambda)=lambda^2-(m_{11}+m_{22})lambda+det(M)#

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 = ((4,0),(0,8))#

with characteristic polynomial

#lambda^2-12lambda+32=0#

with roots #{4,8}# so we have an ellipse.

Being an ellipse there is a canonical representation for it

#((x-x_0)/a)^2+((y-y_0)/b)^2=1#

#x_0,y_0,a,b# can be determined as follows

#4 x^2 + 8 y^2 - 8 x - 28-(b^2(x-x_0)^2+a^2(y-y_0)^2-a^2b^2)=0 forall x in RR#

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

#{a^2 = 8, b^2 = 4, x_0 = 1, y_0 = 0}#

so

#{4 x^2 + 8 y^2 - 8 x - 24=4}equiv{(x-1)^2/8+y^2/4=1}#

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