# How do you find the equation of the ellipse which satisfies the following conditions: the foci are at (+-2,0) and the ellipse passes through (2,-3)?

Apr 30, 2017

Equation of ellipse is ${x}^{2} / 16 + {y}^{2} / 12 = 1$

#### Explanation:

As the foci are $\left(+ 2 , 0\right)$ and $\left(- 2 , 0\right)$, the major axis is aligned with $x$-axis and centre of the ellipse is $\left(0 , 0\right)$. The equation of such an ellipse is of the type

${x}^{2} / {a}^{2} + {y}^{2} / {b}^{2} = 1$, where $2 a$ is major axis and $2 b$ is minor axis and ${b}^{2} = {a}^{2} \left(1 - {e}^{2}\right) = {a}^{2} - {a}^{2} {e}^{2}$. where $e$ is the eccentricity of the ellipse and foci are $\left(a e , 0\right)$ and $\left(- a e , 0\right)$. Here $a e = 2$ and $a > 2$.

Then the equation becomes ${x}^{2} / {a}^{2} + {y}^{2} / \left({a}^{2} \left(1 - {e}^{2}\right)\right) = 1$

As it passes through $\left(2 , - 3\right)$, we have

$\frac{4}{a} ^ 2 + \frac{9}{{a}^{2} - 4} = 1$

or $4 \left({a}^{2} - 4\right) + 9 {a}^{2} = {a}^{2} \left({a}^{2} - 4\right)$ and if ${a}^{2} = k$ it becomes

$4 k - 16 + 9 k = {k}^{2} - 4 k$

or ${k}^{2} - 17 k + 16 = 0$

or $\left(k - 1\right) \left(k - 16\right) = 0$

i.e. $k = {a}^{2} = 1$ or $k = {a}^{2} = 16$

As $a > 2$, we have $a = 4$ and $b = \sqrt{{4}^{2} - {2}^{2}} \sqrt{12}$

and equation of ellipse is ${x}^{2} / 16 + {y}^{2} / 12 = 1$

graph{x^2/16+y^2/12=1 [-10, 10, -5, 5]}