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

2 Answers
Oct 5, 2017

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!

Oct 5, 2017

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]}