How do you find the equation in standard form of an ellipse that passes through the given points: (5, 6), (5, 0), (7, 3), (3, 3)?

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Answer 1 · 2017-07-09T16:17:35

Equation of ellipse is #(x-5)^2/4+(y-3)^2/9=1#

Let the equation of the ellipse be #(x-h)^2/a^2+(y-k)^2/b^2=1#

As it passes through #(5, 6), (5, 0), (7, 3), (3, 3)#, we have

#(5-h)^2/a^2+(6-k)^2/b^2=1# .....(A)

#(5-h)^2/a^2+(0-k)^2/b^2=1# .....(B)

Subtracting (B) from (A) we get

#(6-k)^2/b^2-(0-k)^2/b^2=0# or #(6-k)^2-(0-k)^2=0#

i.e. #36-12k+k^2-k^2=0# or #12k=36# i.e. #k=3#

#(7-h)^2/a^2+(3-k)^2/b^2=1# .....(C)

#(3-h)^2/a^2+(3-k)^2/b^2=1# .....(D)

Subtracting (D) from (C) we get

#(7-h)^2/a^2-(3-h)^2/a^2=0# or #(7-h)^2-(3-h)^2=0#

i.e. #49-14h+h^2-9+6h-h^2=0# or #8h=40# i.e. #h=5#

This reduces the equations (A) and (C) to

#9/b^2=1# i.e. #b^2=9#

and #4/a^2=1# i.e. #a^2=4#

Hence, the equation of ellipse is #(x-5)^2/4+(y-3)^2/9=1#

and ellipse appears as one shown below

graph{((x-5)^2/4+(y-3)^2/9-1)((x-5)^2+(y-6)^2-0.02)((x-5)^2+y^2-0.02)((x-7)^2+(y-3)^2-0.02)((x-3)^2+(y-3)^2-0.02)=0 [-5.71, 14.29, -2.48, 7.52]}