How do you classify the conic #4x^2-y^2-8x+4y-9=0#?

2 Answers
Aug 8, 2017

Answer:

It is a hyperbola.

Explanation:

#4x^2-y^2-8x+4y-9=0#

#hArr4(x^2-2x+1)-4-(y^2+4y+4)+4-9=0#

or #4(x-1)^2-(y+2)^2=9#

or #(x-1)^2/(3/2)^2-(y+2)^2/3^2=1#

As #4x^2-y^2-8x+4y-9=0# can be written as #(x-h)^2/a^2-(y-k)^2/b^2=1#, it is a hyperbola.

graph{4x^2-y^2-8x+4y-9=0 [-9.46, 10.54, -3.76, 6.24]}

Aug 8, 2017

Answer:

The General Cartesian form for a conic section is:

#Ax^2+Bxy+Cy^2+Dx+Ey+F = 0#
One may use the value of #B^2-4AC# to classify it.

Explanation:

The given equation

#4x^2-y^2-8x+4y-9=0#

Fits the General Cartesian form with #A = 4, B, = 0, C = -1, D = -8, E = 4 and F=-9#

Refering to the reference section entitled, Discriminant , we compute the value:

#B^2 -4AC = 0^2-4(4)(-1) = 16#

The value is greater than zero. The list of conditions given by reference tells us that the conic section is a hyperbola.