# How do you write an equation of an ellipse in standard form given vertices (4,-7) and (4,3) and foci (4,-6) and (4,2)?

Jan 9, 2017

${\left(x - 3\right)}^{2} / {3}^{2} - {\left(y + 2\right)}^{2} / {5}^{2} = 1$ The Socratic graph is inserted.

#### Explanation:

graph{(x-3)^2/3^2+(y+2)^2/5^2-1=0 [-20, 20, -10, 10]}

The line joining the foci S(4, 2) and S'(4, -6) is the middle segment of

the the major axis joining the vertices A(4, 3) and A'(4, -7).

It follows that the common midpoint C(4, -2) is the center of the

ellipse, the major axis is along x = 4, in the ($y \uparrow$) and the minor

axis is along y = 2, in the $x \rightarrow$. .

The distance between the vertices

$\forall ' = 2 a = {x}_{A} - {x}_{A} ' = 3 - \left(- 7\right) = 10$.,

giving a = 5. The distance between the foci

$S S ' = 2 a e = 10 e = {x}_{S} - x S ' = 2 - \left(- 6\right) = 8 ,$ giving eccentricity

$e = \frac{4}{5}$.

The semi minor-axis $= b = a \sqrt{1 - {e}^{2}} = 5 \sqrt{1 - \frac{16}{25}} = 3$.

Now, the equation of this ellipse with semi axes a = 5, b = 3, center

at C( 4, -2) and axes parallel to the y and x axes, respectively, is

${\left(x - 3\right)}^{2} / {3}^{2} - {\left(y + 2\right)}^{2} / {5}^{2} = 1$