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

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Answer 1 · 2016-10-30T02:41:51

The standard form is:

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

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.