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

Answer:

#(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#