# How do you write an equation of an ellipse in standard form given eccentricity of 0.5 if the vertices for the ellipse occur at the points ( 0 , 7 ) and ( 0, -9 )?

Oct 19, 2016

Given that the vertices of an ellipse are (0,7) and (0,-9).
We can say that the center of the ellipse is $\left(0 \text{,} \frac{7 - 9}{2}\right) = \left(0 , - 1\right)$ and its major axis is on the y-axis. The length of major axis $2 a = 7 - \left(- 9\right) = 16$
and semi major axis $a = 8$

Now it is given that its eccentricity $\left(e\right) = 0.5$

Again we know

${e}^{2} = \frac{{a}^{2} - {b}^{2}}{a} ^ 2$,where b is the length of semi minor axis.

So ${\left(0.5\right)}^{2} = \frac{{8}^{2} - {b}^{2}}{8} ^ 2$

$\implies \frac{1}{4} = 1 - {b}^{2} / 64$

$\implies {b}^{2} = 64 \times \left(1 - \frac{1}{4}\right) = 48$

$\to b = 4 \sqrt{3}$

So standard equation of the ellipse having center (0,-1) and semi major axis a=8 (along y-axis) and semi minor axis $b = 4 \sqrt{3}$ (parallel to xaxis )is given by

${\left(x - 0\right)}^{2} / {\left(4 \sqrt{3}\right)}^{2} + {\left(y - \left(- 1\right)\right)}^{2} / {8}^{2} = 1$

$\implies {x}^{2} / {\left(4 \sqrt{3}\right)}^{2} + {\left(y + 1\right)}^{2} / {8}^{2} = 1$

$\implies {x}^{2} / 48 + {\left(y + 1\right)}^{2} / 64 = 1$