# How do you find the center, vertices, foci and eccentricity of 25x² + 9y² - 50x - 90y + 25 = 0?

Dec 16, 2015

Memorize the standard form ${\left(x - {x}_{0}\right)}^{2} / {a}_{1}^{2} + {\left(y - {y}_{0}\right)}^{2} / {a}_{2}^{2} = 1$

#### Explanation:

$25 \left({x}^{2} - 2 x\right) + 9 \left({y}^{2} - 10 y\right) = - 25$

$25 \left({x}^{2} - 2 x + 1\right) - 25 + 9 \left({y}^{2} - 10 y + 25\right) - 9 \cdot 25 = - 25$

$25 {\left(x - 1\right)}^{2} + 9 {\left(y - 5\right)}^{2} = 9 \cdot 25$

$\frac{{\left(x - 1\right)}^{2}}{9} + \frac{{\left(y - 5\right)}^{2}}{25} = 1$

${u}^{2} / {b}^{2} + {v}^{2} / {a}^{2} = 1 \mathmr{and} a > b$

x = u + 1 ; y = v + 5; a = 5 ; b = 3

Center is $\left(u , v\right) = 0 R i g h t a r r o w \left(x , y\right) = \left(1 , 5\right)$.

v = 0 Rightarrow u = ±b

u = 0 Rightarrow v = ± a

Vertices are (u, v) in {(0, ± a), (± b, 0)}

(x, y) in {(1, 5 ± a), (1 ± b, 5)} = {(1, 0), (1, 10), (-2, 5), (4, 5)}

Foci are (u, v) = (0, ±c) and c^2 + b^2 = a^2

c = sqrt {25 - 9} = 4 Rightarrow (x,y) = (1, 5 ± 4)

Excentricity is $\frac{c}{a} = \frac{4}{5}$