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

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Answer 1 · 2015-12-16T14:55:11

Memorize the standard form #(x - x_0)^2 / a_1^2 + (y - y_0)^2/a_2^2 = 1#

#25(x^2 - 2x) + 9(y^2 - 10y) = -25#

#25(x^2 - 2x + 1) - 25 + 9(y^2 - 10y + 25) - 9*25 = -25#

#25(x- 1)^2 + 9(y - 5)^2 = 9 * 25#

#frac {(x- 1)^2}{9} + frac{(y - 5)^2}{25} = 1#

#u^2/b^2 + v^2/a^2 = 1 and a > b#

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

Center is #(u, v) = 0 Rightarrow (x,y) = (1,5)#.

#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 #c/a = 4/5#