# 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)?

Jul 9, 2017

Equation of ellipse is ${\left(x - 5\right)}^{2} / 4 + {\left(y - 3\right)}^{2} / 9 = 1$

#### Explanation:

Let the equation of the ellipse be ${\left(x - h\right)}^{2} / {a}^{2} + {\left(y - k\right)}^{2} / {b}^{2} = 1$

As it passes through $\left(5 , 6\right) , \left(5 , 0\right) , \left(7 , 3\right) , \left(3 , 3\right)$, we have

${\left(5 - h\right)}^{2} / {a}^{2} + {\left(6 - k\right)}^{2} / {b}^{2} = 1$ .....(A)

${\left(5 - h\right)}^{2} / {a}^{2} + {\left(0 - k\right)}^{2} / {b}^{2} = 1$ .....(B)

Subtracting (B) from (A) we get

${\left(6 - k\right)}^{2} / {b}^{2} - {\left(0 - k\right)}^{2} / {b}^{2} = 0$ or ${\left(6 - k\right)}^{2} - {\left(0 - k\right)}^{2} = 0$

i.e. $36 - 12 k + {k}^{2} - {k}^{2} = 0$ or $12 k = 36$ i.e. $k = 3$

${\left(7 - h\right)}^{2} / {a}^{2} + {\left(3 - k\right)}^{2} / {b}^{2} = 1$ .....(C)

${\left(3 - h\right)}^{2} / {a}^{2} + {\left(3 - k\right)}^{2} / {b}^{2} = 1$ .....(D)

Subtracting (D) from (C) we get

${\left(7 - h\right)}^{2} / {a}^{2} - {\left(3 - h\right)}^{2} / {a}^{2} = 0$ or ${\left(7 - h\right)}^{2} - {\left(3 - h\right)}^{2} = 0$

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

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

$\frac{9}{b} ^ 2 = 1$ i.e. ${b}^{2} = 9$

and $\frac{4}{a} ^ 2 = 1$ i.e. ${a}^{2} = 4$

Hence, the equation of ellipse is ${\left(x - 5\right)}^{2} / 4 + {\left(y - 3\right)}^{2} / 9 = 1$

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]}