# How do you find the vertex, the directrix, the focus and eccentricity of: (9x^2)/25 + (4y^2)/25 = 1?

##### 1 Answer
Jun 16, 2017

Please see below.

#### Explanation:

This a typical equation of an ellipse centered at origin as it is of the form ${x}^{2} / {a}^{2} + {y}^{2} / {b}^{2} = 1$, as we can write it as

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

As $b > a$, we have major axis $\frac{5}{2} \times 2 = 5$ along $y$ axis and minor axis $\frac{5}{3} \times 2 = \frac{10}{3}$ along $x$ axis.

Vertices, along major axis, are $\left(0 , \pm 2.5\right)$ and co-vertices, along minor axis, are $\left(\pm 1.67 , 0\right)$

and eccentricity given by ${a}^{2} = {b}^{2} \left(1 - {e}^{2}\right)$ i.e. $e = \sqrt{1 - {a}^{2} / {b}^{2}}$ is

$e = \sqrt{1 - \frac{25}{9} \times \frac{4}{25}} = \sqrt{1 - \frac{4}{9}} = \frac{\sqrt{5}}{3} = 0.7454$

Focii, along $y$-axis, are given by $\left(0 , \pm b e\right)$ i.e. $\left(0 , \pm 1.86\right)$

and the two directrix are $y = \pm \frac{b}{e} = \pm 3.354$

graph{(9x^2+4y^2-25)(x^2+(y-1.86)^2-0.005)(x^2+(y+1.86)^2-0.005)=0 [-5.02, 4.98, -2.36, 2.64]}