Question

What kind of conic is defined by the equation `4x^2-y^2+8x-6y+4=0`?

Answer 1

The equation represents a hyperbola.

Explanation:

Rewrite the equation by completing the square for `x` and `y`

`4x^2-y^2+8x-6y+4=0`

`4(x^2+2x)-(y^2+6y)+4=0`

Completing the squares

`4(x^2+2x+(2/2)^2)-(y^2+6y+(6/2)^2)+4-4+9=0`

`4(x^2+2x+1)-(y^2+6y+9)+9=0`

Factorising

`4(x+1)^2-(y+3)^2+9=0`

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

This is the equation of a `"hyperbola with a vertical transverse axis."`

`((y-h)/a)^2-((x-k)/(b))^2=1`

The center is `C=(-1,-3)`

The vertices are `A=(-1,0)` and `A'=(-1,-6)`

graph{((y+3)/3)^2-((x+1)/(3/2))^2=1 [-17.87, 18.18, -13.98, 4.04]}