Question

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

Answer 1

Center: `(0,0)`
Vertices: `(-5,0)`, `(5,0)`, `(0,-4)`, and `(0,4)`

Explanation:

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

`(x-x_c)^2/a^2 + (y-y_c)^2/b^2 = 1`

with the point `(x_c, y_c)` 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 `(0,0)`, or the origin.

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

`a^2 = 25 => a = 5`

`b^2 = 16 => b = 4`

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

Thus, the vertices are located at: `(-5,0)`, `(5,0)`, `(0,-4)`, and `(0,4)`

A graph shows the ellipse:

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