# How do you write an equation of an ellipse in standard form given vertices (-5, 4) and (8, 4) and whose focus is (-4, 4)?

Jul 26, 2016

$\textcolor{b l u e}{{\left(2 x - 3\right)}^{2} / 169 + {\left(y - 4\right)}^{2} / 12 = 1}$

#### Explanation:

Given that the the coordinates of vertices of the ellipse are

$\text{Vertices"->(-5,4) " & } \left(8 , 4\right)$

The ordinates of vertices being same (4)the axis of the ellipse is parallel to x-axis.

$\text{Center} \to \left(\frac{- 5 + 8}{2} , \frac{4 + 4}{2}\right) \to \left(1.5 , 4\right)$

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

color(red)((x-1.5)^2/a^2+(y-4)^2/b^2=1)..... (1)

Now we are to findout a and b.

Again it is also given the coordinate of

$\text{Focus} \to \left(- 4 , 4\right)$

Now a is the distance between center and vertex.

$a = \sqrt{{\left(8 - 1.5\right)}^{2} + {\left(4 - 4\right)}^{2}} = 6.5$

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

$\therefore a e = \sqrt{{\left(- 1.5 - 4\right)}^{2} + {\left(4 - 4\right)}^{2}} = 5.5$

$\implies {a}^{2} {e}^{2} = {5.5}^{2} \implies \frac{{a}^{2} \left({a}^{2} - {b}^{2}\right)}{a} ^ 2 = {5.5}^{2}$

$\implies {6.5}^{2} - {b}^{2} = {5.5}^{2}$

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

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

$\textcolor{b l u e}{{\left(x - 1.5\right)}^{2} / {6.5}^{2} + {\left(y - 4\right)}^{2} / 12 = 1}$

$\textcolor{b l u e}{\implies {\left(2 x - 3\right)}^{2} / 169 + {\left(y - 4\right)}^{2} / 12 = 1}$