# How do you write an equation of an ellipse in standard form given foci are (-2,1) and (-2,5) and vertices are (-2,-1) and (-2,7)?

Jul 28, 2016

$\textcolor{b l u e}{{\left(x + 2\right)}^{2} / 12 + {\left(y - 3\right)}^{2} / 16 = 1}$

#### Explanation:

Given that the the coordinates of vertices of the ellipse are

$\text{Vertices"->(-2,-1) " & } \left(- 2 , 7\right)$

The abscissas of vertices being same (-2)the axis of the ellipse is parallel to y-axis.So the major axis is parallel to y-axis

$\text{Center} \to \left(\frac{- 2 - 2}{2} , \frac{- 1 + 7}{2}\right) \to \left(- 2 , 3\right)$

If a and b are halves of the major axis and minor axis respectively then the standard equation of ellipse may be written as

color(red)((x+2)^2/b^2+(y-3)^2/a^2=1)..... (1)

Now we are to findout a and b.

Again it is also given the coordinate of

$\text{Focii} \to \left(- 2 , 1\right) \mathmr{and} \left(- 2 , 5\right)$

Now a is the distance between center and vertex.

$a = \sqrt{{\left(- 2 + 2\right)}^{2} + {\left(7 - 3\right)}^{2}} = 4$

Now if e represnts eccentricity of the ellipse then

$e = \sqrt{\frac{{a}^{2} - {b}^{2}}{a} ^ 2}$

Now the distance between center and focus is ae

$\therefore a e = \sqrt{{\left(- 2 + 2\right)}^{2} + {\left(3 - 5\right)}^{2}} = 2$

$\implies {a}^{2} {e}^{2} = {2}^{2} \implies \frac{{a}^{2} \left({a}^{2} - {b}^{2}\right)}{a} ^ 2 = {2}^{2}$

$\implies {4}^{2} - {b}^{2} = {2}^{2}$

$\implies {b}^{2} = 12$

Now inserting the value of a and b in equation (1) we get the equation of ellipse as

$\textcolor{b l u e}{{\left(x + 2\right)}^{2} / 12 + {\left(y - 3\right)}^{2} / 16 = 1}$