How do you put 5x^2+18y^2-30x+72y+27=0 in standard form, find the center, the endpoints, vertices, the foci and eccentricity?

Aug 7, 2017

Please observe that the square terms are $5 {x}^{2}$ and $18 {y}^{2}$; this tells us that we should begin the process of finding the center coordinates $\left(h , k\right)$ by adding $5 {h}^{2} + 18 {k}^{2} - 27$ to both sides of the equation:

$5 {x}^{2} - 30 x + 5 {h}^{2} + 18 {y}^{2} + 72 y + 18 {k}^{2} = 5 {h}^{2} + 18 {k}^{2} - 27 \text{ [1]}$

Remove a factor of 5 from the first 3 terms on the left and a factor of 18 from the remaining terms on the left:

$5 \left({x}^{2} - 6 x + {h}^{2}\right) + 18 \left({y}^{2} + 4 y + {k}^{2}\right) = 5 {h}^{2} + 18 {k}^{2} - 27 \text{ [2]}$

Please observe that the trinomial in the first parenthesis fits the pattern ${\left(x - h\right)}^{2} = {x}^{2} - 2 h x + {h}^{2}$, when:

$- 2 h = - 6$

$h = 3$

Substitute ${\left(x - 3\right)}^{2}$ into the trinomial on the left of equation [2] and 3 for h on the right:

$5 {\left(x - 3\right)}^{2} + 18 \left({y}^{2} + 4 y + {k}^{2}\right) = 5 {\left(3\right)}^{2} + 18 {k}^{2} - 27 \text{ [3]}$

Please observe that the trinomial in the second parenthesis fits the pattern ${\left(y - k\right)}^{2} = {y}^{2} - 2 k y + {k}^{2}$, when:

$- 2 k = 4$

$k = - 2$

Substitute ${\left(y - \left(- 2\right)\right)}^{2}$ into the trinomial on the left of equation [3] and -2 for k on the right:

$5 {\left(x - 3\right)}^{2} + 18 {\left(y - \left(- 2\right)\right)}^{2} = 5 {\left(3\right)}^{2} + 18 {\left(- 2\right)}^{2} - 27$

Simplify the right side:

$5 {\left(x - 3\right)}^{2} + 18 {\left(y - \left(- 2\right)\right)}^{2} = 90$

Divide both sides of the equation by 90:

${\left(x - 3\right)}^{2} / 18 + {\left(y - \left(- 2\right)\right)}^{2} / 5 = 1$

Write the denominator as squares:

${\left(x - 3\right)}^{2} / {\left(3 \sqrt{2}\right)}^{2} + {\left(y - \left(- 2\right)\right)}^{2} / {\left(\sqrt{5}\right)}^{2} = 1 \text{ [4]}$

Equation [4] fits the standard Cartesian form for the equation of an ellipse with a horizontal major axis:

${\left(x - h\right)}^{2} / {a}^{2} + {\left(y - k\right)}^{2} / {b}^{2} = 1$

where $h = 3 , k = - 2 , a = 3 \sqrt{2} \mathmr{and} b = \sqrt{5}$. From this we know the following:

The center is:

$\left(h , k\right) = \left(3 , - 2\right)$

The foci are:

$\left(h - \sqrt{{a}^{2} - {b}^{2}} , k\right) = \left(3 - \sqrt{23} , - 2\right) \mathmr{and} \left(h + \sqrt{{a}^{2} - {b}^{2}} , k\right) = \left(3 + \sqrt{23} , - 2\right)$

The vertices are:

$\left(h - a , k\right) = \left(3 - 3 \sqrt{2} , - 2\right) \mathmr{and} \left(h + a , k\right) = \left(3 + 3 \sqrt{2} , - 2\right)$

The endpoints of the minor axis are:

$\left(h , k - b\right) = \left(3 , - 2 - \sqrt{5}\right) \mathmr{and} \left(h , k + b\right) = \left(3 , - 2 + \sqrt{5}\right)$

The eccentricity is:

$\epsilon = \sqrt{1 - {b}^{2} / {a}^{2}} = \sqrt{1 - \frac{5}{18}} = \sqrt{\frac{13}{18}}$