Open Question: How do I find the vertices and foci of this ellipse?

11:54 Publicado por Flechado16

x^2 + 2y^2 + 2x - 20y + 43 = 0
(x^2+2x+1-1) +2 (y^2-10y+25-25) + 43 = 0

(x+1)^2 -1 + 2(y-5)^2 -50 +43 = 0
(x+1)^2 +2(y-5)^2 = 8

divide both sides by 8
(x+1)^2 / 8 + (y-5)^2 /4 = 1

The general form of an ellipse is (x-h)^2 /a^2 + (y-k)^2 / b^2 = 1
(h,k) is the center.
(-1,5) is the center.
The major axis is parallel to the x-axis since a^2=8 > b^2 = 4

a^2=8
b^2=4
c^2=b^2-a^2 = 8-4 = 4
c = +/- 2

Vertices: ( h-a,k) and (h+a,k) = (-1-v8, 5) and (-1+v8, 5)

focit: (h-c, k) and (h+c, k) = (-1-2, 5) and (-1+2,5) = (-3,5) and (1,5)


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