# How do you find center, vertex, and foci of an ellipse 9x^2 + 25y^2 - 36x + 50y - 164 = 0?

Feb 15, 2016

Center of ellipse is $\left(2 , - 1\right)$, vertices are $\left(- 3 , - 1\right)$ and $\left(7 , - 1\right)$ and foci are $\left(- 2 , - 1\right)$ and $\left(6 , - 1\right)$.

#### Explanation:

The equation $9 {x}^{2} + 25 {y}^{2} - 36 x + 50 y - 164 = 0$ can be written as

$9 \left({x}^{2} - 4 x\right) + 25 \left({y}^{2} + 2 y\right) = 164$ or

$9 \left({x}^{2} - 4 x + 4\right) + 25 \left({y}^{2} + 2 y + 1\right) = 164 + 36 + 25$

$9 {\left(x - 2\right)}^{2} + 25 {\left(y + 1\right)}^{2} = 225$

Dividing both sides 225, we get

$\frac{9}{225} {\left(x - 2\right)}^{2} + \frac{25}{225} {\left(y + 1\right)}^{2} = 1$ or ${\left(x - 2\right)}^{2} / {5}^{2} + {\left(y + 1\right)}^{2} / {3}^{2} = 1$

Hence, Center of ellipse is $\left(2 , - 1\right)$ and major axis is $10$ ($2 \cdot 5$) and minor axis is $6$ ($2 \cdot 3$) and vertices are $\left(- 3 , - 1\right)$ and $\left(7 , - 1\right)$.

To find foci, ${c}^{2} = {5}^{2} - {3}^{2} = 16 = {4}^{2}$. Hence foci are $\left(- 2 , - 1\right)$ and $\left(6 , - 1\right)$ (4 units on major axis on either side of center).