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 )?

1 Answer
Jan 28, 2017

Answer:

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.