Question

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)?

Answer 1

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

Explanation:

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]}