How do you find the center and vertices of the ellipse x^2/25+y^2/16=1?

Nov 9, 2017

Center: $\left(0 , 0\right)$
Vertices: $\left(- 5 , 0\right)$, $\left(5 , 0\right)$, $\left(0 , - 4\right)$, and $\left(0 , 4\right)$

Explanation:

The standard form of the equation for an ellipse is given by:

${\left(x - {x}_{c}\right)}^{2} / {a}^{2} + {\left(y - {y}_{c}\right)}^{2} / {b}^{2} = 1$

with the point $\left({x}_{c} , {y}_{c}\right)$ representing the center, the value $a$ representing the horizontal semi-axis length, and the value $b$ representing the vertical semi-axis length.

From inspection of the original equation, we can see that ${x}_{c} = 0$ and ${y}_{c} = 0$, meaning the center of the ellipse is at the point $\left(0 , 0\right)$, or the origin.

Further, we can calculate $a$ and $b$ using the standard form equation:

${a}^{2} = 25 \implies a = 5$

${b}^{2} = 16 \implies b = 4$

From this we know that we can go 5 units to the left and 5 units to the right from the center $\left(0 , 0\right)$ to locate the horizontal vertices, and we can go 4 units above and 4 units below the center $\left(0 , 0\right)$ to locate the vertical vertices.

Thus, the vertices are located at: $\left(- 5 , 0\right)$, $\left(5 , 0\right)$, $\left(0 , - 4\right)$, and $\left(0 , 4\right)$

A graph shows the ellipse:

graph{x^2/25+y^2/16=1 [-10, 10, -6, 6]}