# How do you find the coordinates of the center, foci, the length of the major and minor axis given 3x^2+y^2+18x-2y+4=0?

Jan 10, 2017

Given: ${\left(y - k\right)}^{2} / {a}^{2} + {\left(x - h\right)}^{2} / {b}^{2} = 1 \text{ }$
Center: $\left(h , k\right)$
Foci:$\left(h , k - \sqrt{{a}^{2} - {b}^{2}}\right) \mathmr{and} \left(h , k + \sqrt{{a}^{2} - {b}^{2}}\right)$
major axis = 2a
minor axis = 2b

#### Explanation:

The following are the steps to put the given equation into the form of equation :

Subtract 4 from both sides:

$3 {x}^{2} + {y}^{2} + 18 x - 2 y = - 4 \text{ }$

Group the x terms and the y terms together on the left:

$3 {x}^{2} + 18 x + {y}^{2} - 2 y = - 4 \text{ }$

Because the coefficient of the x^2 term is 3, add #3h^2 to both sides ; make it the 3rd term on the left and the first term on the right:

$3 {x}^{2} + 18 x + 3 {h}^{2} + {y}^{2} - 2 y = 3 {h}^{2} - 4 \text{ }$

Because the coefficient of the y^2 term is 1, add k^2 to both sides; make it the sixth term on the left and the second term on the right:

$3 {x}^{2} + 18 x + 3 {h}^{2} + {y}^{2} - 2 y + {k}^{2} = 3 {h}^{2} + {k}^{2} - 4 \text{ }$

Remove a common factor of 3 from the first 3 terms:

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

Use the pattern for ${\left(x - h\right)}^{2} = {x}^{2} - 2 h x + {h}^{2}$.

Match the "-2hx" term in the pattern with the "6x" term in equation  and write the equation:

$- 2 h x = 6 x$

Solve for h:

$h = - 3$

This means that we can substitute ${\left(x - - 3\right)}^{2}$ for $\left({x}^{2} + 6 x + {h}^{2}\right)$ on the left side of equation  and -3 for h on the right:

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

Use the pattern for ${\left(y - k\right)}^{2} = {y}^{2} - 2 k y + {k}^{2}$.

Match the "-2ky" term in the pattern with the "-2y" term in equation  and write the equation:

$- 2 k y = - 2 y$

Solve for k:

$k = 1$

This means that we can substitute ${\left(y - 1\right)}^{2}$ for ${y}^{2} - 2 y + {k}^{2}$ on the left side of equation  and 1 for k on the right:

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

Simplify the right:

$3 {\left(x - - 3\right)}^{2} + {\left(y - 1\right)}^{2} = 6 \text{ }$

Divide both sides by 6:

${\left(x - - 3\right)}^{2} / 2 + {\left(y - 1\right)}^{2} / 6 = 1 \text{ }$

Swap terms and write the denominators as squares:

${\left(y - 1\right)}^{2} / {\left(\sqrt{6}\right)}^{2} + {\left(x - - 3\right)}^{2} / {\left(\sqrt{2}\right)}^{2} = 1 \text{ }$

We have the form of equation 

$h = - 3$
$k = 1$
$a = \sqrt{6}$
$b = \sqrt{2}$
$\sqrt{{a}^{2} - {b}^{2}} = \sqrt{6 - 2} = \sqrt{4} = 2$

Center: $\left(- 3 , 1\right)$
Foci: $\left(- 3 , 1 - 2\right) \mathmr{and} \left(- 3 , 1 + 2\right) = \left(- 3 , - 1\right) \mathmr{and} \left(- 3 , 3\right)$
Major axis: $2 \sqrt{6}$
Minor axis: $2 \sqrt{2}$