# How do you find all the critical points to graph x^2/9 + y^2/25 = 1 including vertices, foci and asymptotes?

Jun 15, 2016

Vertices are $\left(- 5 , 0\right)$ and $\left(5 , 0\right)$ and focii are $\left(0 , - 4\right)$ and $\left(0 , 4\right)$.

#### Explanation:

The given equation, ${x}^{2} / 9 + {y}^{2} / 25 = 1$ or ${x}^{2} / {3}^{2} + {y}^{2} / {5}^{2} = 1$,

is clearly the standard form of equation of an ellipse with center at $\left(0 , 0\right)$, major axis as $5 \times 2 = 10$ (at $y$-axis) and minor axis as $3 \times 2 = 6$ (at $x$-axis). Clearly vertices are $\left(- 5 , 0\right)$ and $\left(5 , 0\right)$.

Eccentricity is $e = \sqrt{1 - {3}^{2} / {5}^{2}} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} = 0.8$.

And hence focii are $\left(0 , - b e\right)$ and $\left(0 , b e\right)$ i.e. $\left(0 , - 4\right)$ and $\left(0 , 4\right)$.

As it is an ellipse question of asymptote does not arise

The graph appears as below.

graph{x^2/9+y^2/25=1 [-10, 10, -5, 5]}