Answer
$\frac{x^2}{9}+\frac{y^2}{25} = 1$
Work Step by Step
We can write the general form of an ellipse:
$\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2} = 1$
Since the center is the origin, we can rewrite the equation of the ellipse as follows:
$\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$
We can use the two data pairs to make two equations:
equation 1:
$\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$
$\frac{(1)^2}{a^2}+\frac{(-10\sqrt{2}/3)^2}{b^2} = 1$
$\frac{1}{a^2}+\frac{200}{9b^2} = 1$
$\frac{-4}{a^2}+\frac{-800}{9b^2} = -4$
equation 2:
$\frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$
$\frac{(-2)^2}{a^2}+\frac{(5\sqrt{5}/3)^2}{b^2} = 1$
$\frac{4}{a^2}+\frac{125}{9b^2} = 1$
We can add both sides of equation 1 and equation 2:
$\frac{-800}{9b^2}+\frac{125}{9b^2} = -4+1$
$\frac{-675}{9b^2} = -3$
$b^2 = \frac{675}{27}$
$b^2 = \frac{225}{9}$
$b^2 = \frac{75}{3}$
$b^2 = 25$
$b = 5$
We can use equation 1 to find $a$:
$\frac{1}{a^2}+\frac{200}{9b^2} = 1$
$\frac{1}{a^2}+\frac{200}{9(5)^2} = 1$
$\frac{1}{a^2}+\frac{8}{9} = 1$
$\frac{1}{a^2}=\frac{1}{9}$
$a^2 = 9$
$a = 3$
We can write the equation of the ellipse:
$\frac{x^2}{9}+\frac{y^2}{25} = 1$
