How do you solve the standard form of an ellipse given Foci: (0, -3) , (0, 3) Vertices: (0, -4), (0, 4)?

1 Answer
Oct 24, 2016

#x^2/7 + y^2/16 = 1#

Explanation:

#f_1: (0, -3)#

#f_2: (0, 3)#

#V_1: (0, -4)#

#V_2: (0, 4)#


Distance #D_f# between foci #=> 2c#
Distance #D_V# between vertices #=> 2a#

#D = sqrt((x_1 - x_2)^2 + (y_1 - y_2)^2)#


#D_f = sqrt((0 - 0)^2 + (-3 - 3)^2)#
#=> D_f = sqrt(-6^2) = 6#
#=> c = 3#

#D_V = sqrt((0 - 0)^2 + (-4 - 4)^2)#
#=> D_V = sqrt(-8^2) = 8#
#=> a = 4#


#c^2 = a^2 - b^2#
#=> b^2 = a^2 - c^2#

#b^2 = 4^2 - 3^2#
#b^2 = 16 - 9 = 7#

#=> b = sqrt7#


Midpoint #M# between the foci and/or vertices #=># center #C: (h, k)# of the ellipse

#M: ((x_1 + x_2)/2,(y_1 + y_2)/2)#

Using either the foci or the vertices, we get

#C: (0,0)#


Note that between the foci and between the vertices, only the y-coordinate changes. This means the ellipse has a vertical major axis

Standard Equation of an ellipse with vertical major axis

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

#(x - 0)^2/(sqrt7)^2 + (y - 0)^2/(4^2) = 1#

#=> x^2/7 + y^2/16 = 1#