# How do you put 6x^2+5y^2-24x+20y+14=0 in standard form, find the center, the endpoints, vertices, the foci and eccentricity?

Mar 15, 2017

Standard forms for the equation of an ellipse are ${\left(x - h\right)}^{2} / {a}^{2} + {\left(y - k\right)}^{2} / {b}^{2} = 1$ or (y-k)^2/a^2+(x-h)^2/b^2=1; a > b

#### Explanation:

Reference for an ellipse
Given: $6 {x}^{2} + 5 {y}^{2} - 24 x + 20 y + 14 = 0$

Add $6 {h}^{2} + 5 {k}^{2} - 14$ to both sides of the equation:

$6 {x}^{2} - 24 x + 6 {h}^{2} + 5 {y}^{2} + 20 y + 5 {k}^{2} = 6 {h}^{2} + 5 {k}^{2} - 14$

Remove a common factor of 6 from the first 3 terms and a common factor of 5 from the next 3 terms:

$6 \left({x}^{2} - 4 x + {h}^{2}\right) + 5 \left({y}^{2} + 4 y + {k}^{2}\right) = 6 {h}^{2} + 5 {k}^{2} - 14 \text{ [1]}$

To find the value of h, set the middle term of the right side of the pattern ${\left(x - h\right)}^{2} = {x}^{2} - 2 h x + {h}^{2}$ equal to the middle term in the first parenthesis in equation [1]:

$- 2 h x = - 4 x$

$h = 2$

Substitute the left side of the pattern into equation [1]:

$6 {\left(x - h\right)}^{2} + 5 \left({y}^{2} + 4 y + {k}^{2}\right) = 6 {h}^{2} + 5 {k}^{2} - 14 \text{ [2]}$

Substitute 2 for h everywhere in equation [2]:

$6 {\left(x - 2\right)}^{2} + 5 \left({y}^{2} + 4 y + {k}^{2}\right) = 6 {\left(2\right)}^{2} + 5 {k}^{2} - 14 \text{ [3]}$

To find the value of k, set the middle term of the right side of the pattern ${\left(y - k\right)}^{2} = {y}^{2} - 2 k y + {k}^{2}$ equal to the middle term in the second parenthesis in equation [3]:

$- 2 k y = 4 y$

$k = - 2$

Substitute the left side of the pattern into equation [3]:

$6 {\left(x - 2\right)}^{2} + 5 {\left(y - k\right)}^{2} = 6 {\left(2\right)}^{2} + 5 {k}^{2} - 14 \text{ [4]}$

Substitute -2 for k everywhere in equation [4]:

$6 {\left(x - 2\right)}^{2} + 5 {\left(y - - 2\right)}^{2} = 6 {\left(2\right)}^{2} + 5 {\left(- 2\right)}^{2} - 14 \text{ [5]}$

Simplify the right side of equation [5]:

$6 {\left(x - 2\right)}^{2} + 5 {\left(y - - 2\right)}^{2} = 30 \text{ [6]}$

Divide both sides of the equation by 30:

${\left(x - 2\right)}^{2} / 5 + {\left(y - - 2\right)}^{2} / 6 = 1 \text{ [7]}$

Make the denominators squares and swap terms:

${\left(y - - 2\right)}^{2} / {\left(\sqrt{6}\right)}^{2} + {\left(x - 2\right)}^{2} / {\left(\sqrt{5}\right)}^{2} = 1 \text{ [8]}$

Equation [8] is the standard form where, $h = 2 , k = - 2 , a = \sqrt{6} , \mathmr{and} b = \sqrt{5}$

From the reference:

$c = \sqrt{{a}^{2} - {b}^{2}}$

$c = \sqrt{6 - 5}$

$c = 1$

The center is at $\left(h , k\right)$:

$\left(2 , - 2\right)$

The vertices are at (h, k-a) and (h,k+a):

$\left(2 , - 2 - \sqrt{6}\right) \mathmr{and} \left(2 , - 2 + \sqrt{6}\right)$

The foci are at: (h, k-c) and (h,k+c):

$\left(2 , - 2 - 1\right) \mathmr{and} \left(2 , - 2 + 1\right)$

$\left(2 , - 3\right) \mathmr{and} \left(2 , - 1\right)$

The co-vertices are at $\left(h - b , k\right) \mathmr{and} \left(h + b , k\right)$:

$\left(2 - \sqrt{5} , - 2\right) \mathmr{and} \left(2 + \sqrt{5} , - 2\right)$

The eccentricity is $\frac{c}{a} = \frac{1}{\sqrt{6}} = \frac{\sqrt{6}}{6}$