How do you write an equation of an ellipse in standard form given Foci (0,3),(0,-3) Passing through points (-1,2√2)?

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Answer 1 · 2018-07-17T13:05:20

Equation of vertical ellipse is $x^2/3+y^2/12=1;$

Focii are $f_1 ( 0,3) , f_2 (0,-3) :. f_1 f_2=6 :. c=6/2=3$

Point on ellipse is $P(-1,2 sqrt2)$

Distance between two points:

$D= sqrt((x_1-x_2)^2+(y_1-y_2)^2)$

$d_1=P f_1=sqrt((-1-0)^2+(2sqrt2-3)^2)~~1.0146$

$d_2=P f_2=sqrt((-1-0)^2+(2sqrt2+3)^2)~~5.9135$

$:. d_1+d_2~~ 6.928 = 2 a :.a=6.928/2~~3.464 = 2 sqrt3$

$b^2=a^2-c^2 :. b^2 =12-9 :. b^2=3 :. b=sqrt 3$

Equation of vertical ellipse is $x^2/b^2+y^2/a^2=1; a>b$

Hence equation of vertical ellipse is $x^2/3+y^2/12=1;$

graph{x^2/3 + y^2/12= 1 [-10, 10, -5, 5]} [Ans]