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

Oct 5, 2017

$4 {x}^{2} + 8 x + 3 {y}^{2} + 12 y + 1 = 0$
$\implies {\left(2 x + 1\right)}^{2} + {\left(\sqrt{3} y + 2 \sqrt{3}\right)}^{2} = 12$

#### Explanation:

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

$4 {x}^{2} + 8 x + 3 {y}^{2} + 12 y + 1 = 0$

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

$4 {x}^{2} + 8 x + 1 + 3 {y}^{2} + 12 y + 1 = 1$

${\left(2 x + 1\right)}^{2} + 3 {y}^{2} + 12 y + 1 = 1$

Now for $y$:

${\left(2 x + 1\right)}^{2} + 3 {y}^{2} + 12 y + 12 + 1 = 1 + 12$

${\left(2 x + 1\right)}^{2} + {\left(\sqrt{3} y + 2 \sqrt{3}\right)}^{2} + 1 = 13$

${\left(2 x + 1\right)}^{2} + {\left(\sqrt{3} y + 2 \sqrt{3}\right)}^{2} = 12$

Hope this helps!

Oct 5, 2017

#### Explanation:

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

$4 \left({x}^{2} + 2 x\right) + 3 \left({y}^{2} - 4 y\right) + 1 = 0$

or $4 \left({x}^{2} + 2 x + 1\right) + 3 \left({y}^{2} - 4 y + 4\right) + 1 - 16 = 0$

or $4 {\left(x + 1\right)}^{2} + 3 {\left(y - 2\right)}^{2} = 15$

or ${\left(x + 1\right)}^{2} / \left(\frac{15}{4}\right) + {\left(y - 2\right)}^{2} / \left(\frac{15}{3}\right) = 1$

or ${\left(x + 1\right)}^{2} / {\left(\frac{\sqrt{15}}{2}\right)}^{2} + {\left(y - 2\right)}^{2} / \left(\sqrt{5}\right) = 1$

Hence, it is an equation of an ellipse, whose major axis is vertical (parallel to $y$-axis) and is $2 \sqrt{5}$ and minor axis is horizontal (parallel to $x$-axis).

Its center is $\left(- 1 , 2\right)$ and eccentricity is $\sqrt{1 - \frac{3}{4}} = \frac{1}{2}$ and focii are $\left(- 1 , 2 \pm \sqrt{5}\right)$, (i.e. $2 \pm a e$, 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]}