How do you find the standard form of the equation of the ellipse given the properties foci #(0,+-5)#, vertices #(0, +-8)#?

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Answer 1 · 2018-07-28T11:51:49

The equation of the ellipse is #y^2/64+x^2/39=1#

The equation of an ellipse with major vertical axis is

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

The center( symmetric wrt the foci and the vertices) of the ellipse is

#C=(h,k)=(0,0)#

Therefore,

#a=8#

#c=5#

#b^2=(a^2-c^2)=(64-25)=sqrt39#

The equation of the ellipse is

#y^2/64+x^2/39=1#

graph{(y^2/64+x^2/39-1)=0 [-17.3, 18.75, -8.67, 9.35]}

Answer 2 · 2018-07-28T12:04:31

The standard equation of vertical ellipse is # x^2/39+y^2/64=1#

The vertices and foci are on the y axis at

#V(0, 8), V'(0, -8) and F(0, 5) , F'(0, -5)#

Semi major axis is #a=8# and focus #c=5# from the center

#(0,0)#. This is vertical ellipsce of which the equation is

#x^2/b^2+y^2/a^2=1 or x^2/b^2+y^2/8^2=1#

#c=5# is the distance from the center to a focus. The relation of

#c, a, b# is # c^2 = a^2 - b^2:. 5^2=8^2-b^2 # or

#b^2=64-25=39 :. b= sqrt 39 ~~6.245 #, therefore, the equation

of vertical ellipse is # x^2/39+y^2/64=1#

graph{x^2/39+y^2/64=1 [-20, 20, -10, 10]} [Ans]