# How do you find the coordinates of the center, foci, lengths of the major and minor axes given x^2+25y^2-8x+100y+91=0?

Dec 30, 2016

$C : \left(4 , - 2\right)$
$f : \left(4 \pm 2 \sqrt{6} , - 2\right)$
Major axis length: $10$
Minor axis length: $2$

#### Explanation:

${x}^{2} + 25 {y}^{2} - 8 x + 100 y + 91 = 0$

Group the terms with the same variables and simplify

$\implies \left({x}^{2} - 8 x\right) + \left(25 {y}^{2} + 100 y\right) + 91 = 0$

$\implies \left({x}^{2} - 8 x\right) + 25 \left({y}^{2} + 4 y\right) + 91 = 0$

Add a third term to the grouped terms such that it will be a perfect square trinomial. Then, subtract the same value on the same side of the equation or add the same value to the other side of the equation. This will ensure that the equality is maintained.

${\left(x + a\right)}^{2} = {x}^{2} + 2 a + {a}^{2}$

$\implies \left({x}^{2} - 8 x + 16\right) + 25 \left({y}^{2} + 4 y + 4\right) + 91 - 16 - 25 \left(4\right) = 0$

$\implies {\left(x - 4\right)}^{2} + 25 {\left(y + 2\right)}^{2} - 25 = 0$

$\implies {\left(x - 4\right)}^{2} + 25 {\left(y + 2\right)}^{2} = 25$

Finally, divide the entire equation by the constant on the right hand side.

$\implies \frac{{\left(x - 4\right)}^{2} + 25 {\left(y + 2\right)}^{2} = 25}{25}$

$\implies {\left(x - 4\right)}^{2} / 25 + {\left(y + 2\right)}^{2} / 1 = 1$

$C : \left(h , k\right)$
$\implies C : \left(4 , - 2\right)$

The larger denominator is with $x$, the major axis is horizontal

$f : \left(h \pm c , k\right)$

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

${c}^{2} = 25 - 1$

${c}^{2} = 24$

$c = \sqrt{24}$

$c = 2 \sqrt{6}$

$\implies f : \left(4 \pm 2 \sqrt{6} , - 2\right)$

Major axis: $2 a = 10$
Minor axis: $2 b = 2$