# How do you find the center, vertices, and foci of an ellipse #(x-3)^2 / 16 + (y+2)^2 / 9=1#?

##### 3 Answers

Please see the explanation.

#### Explanation:

The standard form for the equation of an ellipse is:

The center is:

The vertices on the major axis are:

The vertices on the minor axis are:

The foci are:

To put the given equation in standard form, change the + 2 to - -2 and write the denominators as squares:

The center is:

The vertices on the major axis are:

The vertices on the minor axis are:

Evaluate:

The foci are:

The center is

The vertices are

The foci are F

#### Explanation:

We compare this equation to

The center is

The vertices are A

and

B

To calculate the foci, we need c

The foci are F

and F'

graph{(x-3)^2/16+(y+2)^2/9=1 [-8.56, 11.44, -8.316, 1.685]}

Centre is (3, -2), focii are

#### Explanation:

Standard equation of an ellipse centered at (h,k) is

The foci of this ellipse are at (c+h, k) and (-c+h, k). The vertices on horizontal axis would be at (-a+h,k) and (a+h,k), where

Comparing the given equation with the standard one, it is seen that a=4, b=3, c=

Hence Centre is (3, -2), focii are