How do you write an equation of an ellipse in standard form given center at origin and passes through (√6, 2) and (-3, √2)?

1 Answer
Oct 30, 2016

The standard form is:

(x - 0)^2/(sqrt(12))^2 + (y - 0)^2/(sqrt(8))^2 = 1

Explanation:

The standard form of an equation of an ellipse with an arbitrary center (h, k) is:

(x - h)^2/a^2 + (y - k)^2/b^2 = 1

The standard form equation for an ellipse with its center at the origin is:

(x - 0)^2/a^2 + (y - 0)^2/b^2 = 1

Write two equations using the above form and the two given points:

(sqrt(6)- 0)^2/a^2 + (2- 0)^2/b^2 = 1 [1]
(-3- 0)^2/a^2 + (sqrt(2)- 0)^2/b^2 = 1 [2]

Do the multiplication implied by the squares:

6/a^2 + 4/b^2 = 1 [3]
9/a^2 + 2/b^2 = 1[4]

Let u = 1/a^2 and let v = 1/b^2

6u + 4v = 1 [5]
9u + 2v = 1 [6]

Multiply equation [6] by -2 and add to equation [5]

6u - 18u + 4v - 4v = 1 - 2

-12u = -1

u = 1/12

Substitute 1/12 for u in equation [6]

9/12 + 2v = 1

2v = 3/12

v = 3/24 = 1/8

a = sqrt(12) and b = sqrt(8)

The standard form is:

(x - 0)^2/(sqrt(12))^2 + (y - 0)^2/(sqrt(8))^2 = 1

check:

(sqrt(6) - 0)^2/(sqrt(12))^2 + (2 - 0)^2/(sqrt(8))^2 = 1
(-3 - 0)^2/(sqrt(12))^2 + (sqrt(2) - 0)^2/(sqrt(8))^2 = 1

6/12 + 4/8 = 1
9/12 + 2/8 = 1

1 = 1
1 = 1

This checks.