Question

How do you convert this equation `4x ^ { 2} + 8x + 3y ^ { 2} - 12y + 1= 0` to its graphable form?

Answer 1

By completing the square.

`4x^2+8x+3y^2+12y+1=0`
`=> (2x+1)^2+(sqrt(3)y+2sqrt(3))^2=12`

Explanation:

This equation can be made into its graphable form by completing the square.

`4x^2+8x+3y^2+12y+1=0`

Let's complete the square for `x` first (assuming you know how to complete the square):

`4x^2+8x+1 +3y^2+12y+1=1`

`(2x+1)^2+3y^2+12y+1=1`

Now for `y`:

`(2x+1)^2+3y^2+12y+12+1=1+12`

`(2x+1)^2+(sqrt(3)y+2sqrt(3))^2+1=13`

`(2x+1)^2+(sqrt(3)y+2sqrt(3))^2=12`

And there's your graphable form.

Hope this helps!

Answer 2

Please see below.

Explanation:

`4x^2+8x+3y^2-12y+1=0` can be written as

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

or `4(x^2+2x+1)+3(y^2-4y+4)+1-16=0`

or `4(x+1)^2+3(y-2)^2=15`

or `(x+1)^2/(15/4)+(y-2)^2/(15/3)=1`

or `(x+1)^2/(sqrt15/2)^2+(y-2)^2/(sqrt5)=1`

Hence, it is an equation of an ellipse, whose major axis is vertical (parallel to `y`-axis) and is `2sqrt5` and minor axis is horizontal (parallel to `x`-axis).

Its center is `(-1,2)` and eccentricity is `sqrt(1-3/4)=1/2` and focii are `(-1,2+-sqrt5)`, (i.e. `2+-ae`, where `a` is half major axis and `e` is eccentricity). Its graph appears as shown below.

graph{(4x^2+8x+3y^2-12y+1)((x+1)^2+(y-2)^2-0.01)((x+1)^2+(y-2+sqrt(5/4))^2-0.01)((x+1)^2+(y-2-sqrt(5/4))^2-0.01)=0 [-7, 6, -1, 5]}