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]}