Question

How do you write a standard form equation for the hyperbola with `y^2 - 4y - 4x^2 + 8x = 4`?

Answer 1

The equation is `(y-2)^2/4-(x-1)^2/1=1`

Explanation:

The general equation for a hyperbola is

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

or

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

Here, we have

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

Complete the square

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

`(y-2)^2-4(x-1)^2=4`

Dividing by `4`

`(y-2)^2/4-4(x-1)^2/4=1`

`(y-2)^2/4-(x-1)^2/1=1`

The center is `C=(1,2)` and the vertices are `A=(1,0)` and `A'=(1,4)`

graph{((y-2)^2/4-(x-1)^2/1-1)=0 [-20.43, 20.11, -9, 11.28]}

Answer 2

Standard form of vertical hyperbola equation is
`(y-2)^2/2^2 -(x-1)^2/(1^2) = 1`

Explanation:

`y^2-4 y -4 x^2+8 x= 4` or

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

`(y-2)^2 -4(x-1)^2 = 4` or

`(y-2)^2/4 -(cancel4(x-1)^2)/cancel4 = 1` or

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

The standard form of equation of vertical hyperbola is

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

Therefore, the standard form of vertical hyperbola equation

is `(y-2)^2/2^2 -(x-1)^2/(1^2) = 1`

graph{y^2-4 y -4 x^2+ 8 x -4=0 [-40, 40, -20, 20]} [Ans]