How do you write an equation of an ellipse in standard form given Center at (-3,1); Vertex at (-3,3); Focus at (-3,0)?

1 Answer
Dec 29, 2015

(x + 3)^2/4 + (y - 1)^2/3 = 1

Explanation:

When the major axis is horizontal, the standard equation of an ellipse is

(x - h)^2/a^2 + (y - k)^2/b^2 = 1

where
C: (h, k)
V: (h +- a, k)

c^2 = a^2 - b^2

f: (h +- c, k)


On the other hand, when the major axis is vertical, the standard equation of an ellipse is

(x - h)^2/b^2 + (y - k)^2/a^2 = 1

where
C: (h, k)
V: (h, k +- a)

c^2 = a^2 - b^2

f: (h, k +- c)


In the given,
C: (-3, 1)
V_1: (-3, 3)
f_1: (-3, 0)

Since the x-coordinate of the points are constant, we can say that the major axis is vertical.

C: (h, k)
C: (-3, 1)

=> h = -3
=> k = 1


V: (h, k +- a)

V_1: (-3, 3)

=> 3 = k +- a
=> 3 = 1 +- a

=> 3 = 1 + a
=> 3 = 1 - a

=> 2 = a
=> -2 = a

Since a is the distance between the center and the vertex, we take a to be positive

=> a = 2


f: (h, k +- c)

=> 0 = 1 +- c
=> c = 1
=> c = -1

Similarly for c, since it is the distance between the focus and the center, we take c to be positive

=> c = 1


c^2 = a^2 - b^2
1^2 = 2^2 - b^2

=> 1 = 4 - b^2
=> b^2 = 3
=> b = +-sqrt3

Again, since b is the distance between the center and the end of the minor axis (not sure what it was called), we take b to be positive

=> b = sqrt3


Hence, the equation of the ellipse is

(x - -3)^2/2^2 + (y - 1)^2/(sqrt3)^2 = 1

=> (x + 3)^2/4 + (y - 1)^2/3 = 1