# How do you put 12x^2+3y^2-30y+39=9 in standard form, find the center, the endpoints, vertices, the foci and eccentricity?

Dec 30, 2017

(4x^2)/15 + (y-5)^2/15 = 1; " center": (0, 5)
vertices: $\left(0 , 5 + \sqrt{15}\right) , \left(0 , 5 - \sqrt{15}\right)$
endpoints (co-vertices): $\left(- \frac{\sqrt{15}}{2} , 5\right) , \left(\frac{\sqrt{15}}{2} , 5\right)$
foci: $\left(0 , 5 + \frac{3 \sqrt{5}}{2}\right) , \left(0 , 5 - \frac{3 \sqrt{5}}{2}\right)$; $e = \frac{\sqrt{3}}{2}$

#### Explanation:

Given: $12 {x}^{2} + 3 {y}^{2} - 30 y + 39 = 9$

To Find the standard form of the equation , use the completing of the square technique :

Group both the $x$-terms and the $y$-terms together on the left-side of the equation and all the other constants on the right side:
$12 {x}^{2} + \left(3 {y}^{2} - 30 y\right) = - 30$

Factor if needed: $\text{ } 12 {x}^{2} + 3 \left({y}^{2} - 10 y\right) = - 30$

We only need to use completing of the square with the $y$-terms. Half the $- 10 y \text{ term "-> -5 " and add the } 3 {\left(- 5\right)}^{2}$ to the right side that was added to the left side when we completed the square:

$\text{ } 12 {x}^{2} + 3 {\left(y - 5\right)}^{2} = - 30 + 3 {\left(5\right)}^{2}$

$\text{ } 12 {x}^{2} + 3 {\left(y - 5\right)}^{2} = 45$

Divide everything by $45$ to make the equation $= 1$:

Standard form of the equation:

$\text{ "(4x^2)/15 + ((y-5)^2)/15 = 1 " or } {x}^{2} / {\left(\frac{\sqrt{15}}{2}\right)}^{2} + {\left(y - 5\right)}^{2} / \left({\sqrt{15}}^{2}\right) = 1$

Because we have a "sum" in the equation, we have an ellipse . In an ellipse $a$ is always the major axis (longest). This means we have a Vertical Major Axis ellipse: ${\left(x - h\right)}^{2} / {b}^{2} + {\left(y - k\right)}^{2} / {a}^{2} = 1$

Where $\text{ **center** } = \left(h , k\right) = \left(0 , 5\right)$

$a = \sqrt{15} , \text{ } b = \frac{\sqrt{15}}{2}$
$c = \sqrt{{a}^{2} - {b}^{2}} = \sqrt{15 - \frac{15}{4}} = \sqrt{\frac{45}{4}} = \frac{3 \sqrt{5}}{2}$

Vertices: $\left(h , k + a\right) , \left(h , k - a\right) : \left(0 , 5 + \sqrt{15}\right) , \left(0 , 5 - \sqrt{15}\right)$

Endpoints (co-vertices): $\left(h - b , k\right) , \left(h + b , k\right) : \left(- \frac{\sqrt{15}}{2} , 5\right) , \left(\frac{\sqrt{15}}{2} , 5\right)$

Foci: $\left(h , k + c\right) , \left(h , k - c\right) : \left(0 , 5 + \frac{3 \sqrt{5}}{2}\right) , \left(0 , 5 - \frac{3 \sqrt{5}}{2}\right)$

Eccentricity:
$\varepsilon = \frac{c}{a} = \frac{\frac{3 \sqrt{5}}{2}}{\sqrt{15}} = \frac{3 \sqrt{5}}{2} \cdot \frac{1}{\sqrt{15}} \cdot \frac{\sqrt{15}}{\sqrt{15}} = \frac{3 \sqrt{75}}{30} = \frac{\sqrt{75}}{10}$

$\varepsilon = \frac{\sqrt{3}}{2}$