Question

How do you write an equation of an ellipse in standard form passing through the center at the origin that passes through the points (1, (-10√2)/3 ) and (-2, (5√5)/3 )?

Answer 1

The standard form of the equation of an ellipse is:
`(x-h)^2/a^2+(y-k)^2/b^2=1" [1]"`
where `(h,k)` is the center.

Explanation:

We are given that the center is the origin, `(0,0)`, therefore, we can substitute 0 for h and 0 for k into equation [1] to give us equation [2]:

`(x-0)^2/a^2+(y-0)^2/b^2=1" [2]"`

Use the two given points and equation [2] to write two equations:

`(1-0)^2/a^2 + ((-10sqrt2)/3-0)^2/b^2 = 1" [3]"`

`(-2-0)^2/a^2 + ((5sqrt5)/3-0)^2/b^2 = 1" [4]"`

Let `u = 1/a^2`, let `v = 1/b^2` and square the numerators:

`u + (200/9)v = 1" [5]"`

`4u + (125/9)v = 1" [6]"`

Convert to an augmented matrix :

#[(1, 200/9,|,1), (4, 125/9,|,1) ]#

Perform Elementary row operations :

`R_2-3R_1toR_2`

#[(1, 200/9,|,1), (0, -675/9,|,-3) ]#

`-9R_2/675`

#[(1, 200/9,|,1), (0, 1,|,1/25) ]#

`R_1-200/9R_2toR_1`

#[(1, 0,|,1/9), (0, 1,|,1/25) ]#

`u = 1/9 and v = 1/25`

`1/a^2 = 1/9 and 1/b^2 = 1/25`

`a = 3 and b = 5`

Substitute this into equation [2]:

`(x-0)^2/3^2+(y-0)^2/5^2=1" [7]"`

Equation [7] is the answer.