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

Jul 17, 2018

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

#### Explanation:

Focii are ${f}_{1} \left(0 , 3\right) , {f}_{2} \left(0 , - 3\right) \therefore {f}_{1} {f}_{2} = 6 \therefore c = \frac{6}{2} = 3$

Point on ellipse is $P \left(- 1 , 2 \sqrt{2}\right)$

Distance between two points:

$D = \sqrt{{\left({x}_{1} - {x}_{2}\right)}^{2} + {\left({y}_{1} - {y}_{2}\right)}^{2}}$

${d}_{1} = P {f}_{1} = \sqrt{{\left(- 1 - 0\right)}^{2} + {\left(2 \sqrt{2} - 3\right)}^{2}} \approx 1.0146$

${d}_{2} = P {f}_{2} = \sqrt{{\left(- 1 - 0\right)}^{2} + {\left(2 \sqrt{2} + 3\right)}^{2}} \approx 5.9135$

$\therefore {d}_{1} + {d}_{2} \approx 6.928 = 2 a \therefore a = \frac{6.928}{2} \approx 3.464 = 2 \sqrt{3}$

${b}^{2} = {a}^{2} - {c}^{2} \therefore {b}^{2} = 12 - 9 \therefore {b}^{2} = 3 \therefore b = \sqrt{3}$

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]