How do you find the equation of an ellipse with vertices #(0,+-5)# and passes through the point (4,2)?

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Answer 1 · 2017-10-10T17:40:33

Because the vertices are vertically oriented, we use the general Cartesian form:

#(y-k)^2/a^2+(x-h)^2/b^2=1; a > b" [1]"#

where #(h,k)# is the center.

The general form for vertically oriented vertices are:

#(h, k - a)# and #(h,k-a)#

These general forms and the given vertices #(0,-5) and (0,5)# allow us to write 3 equations that can be used to find the values of #h, k, and a#:

#h = 0#

#k - a = -5#

#k+ a = 5#

#2k = 0#

#k = 0#

#a = 5#

Substitute these values into equation [1]:

#(y-0)^2/5^2+(x-0)^2/b^2=1; a > b" [2]"#

Substitute the point #(4,2)# into equation [2] and solve for b:

#(2-0)^2/5^2+(4-0)^2/b^2=1; a > b#

#4/25+16/b^2 = 1#

#16/b^2 = 21/25#

#b^2 = (16(25))/21#

#b = (20sqrt21)/21#

#(y-0)^2/5^2+(x-0)^2/((20sqrt21)/21)^2=1" [3]"#

Here is a graph of equation [3] and the three points:

www.desmos.com/calculator