How do you find center, vertex, and foci of an ellipse #x^2 + 4y^2 = 1#?

1 Answer
Nov 15, 2015

Answer:

#C: (0, 0)#
#V: (+- 1, 0)#
#f: (+-sqrt3/2, 0)#

Explanation:

The standard equation of an ellipse is either in the form

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

or

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

where #a > b#


In the given equation

#x^2 + 4y^2 = 1#

This is equivalent to

#(x - 0)^2/1^2 + (y - 0)^2/(1/2)^2 = 1#


Our center is at #(h, k)#

#C: (0, 0)#

Since #a# is under #x#, the major axis is horizontal. The vertex is at

#V: (h +- a, k)#

#V_1: (0 + 1, 0) => (1, 0)#

#V_2: (0 - 1, 0) => (-1, 0)#

Meanwhile, the foci are #c# units from the center.
Where #c^2 = a^2 - b^2#

#=> c^2 = 1^2 - (1/2)^2#

#=> c^2 = 1 - 1/4#

#=> c^2 = (4 - 1)/4#

#=> c^2 = 3/4#

#=> c = sqrt3/2#

#f: (h +- c, k)#

#f_1: (0 + sqrt3/2, 0) => (sqrt3/2, 0)#

#f_2: (0 - sqrt3/2, 0) => (-sqrt3/2, 0)#