Question

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

Answer 1

Please see the explanation.

Explanation:

The standard form for the equation of an ellipse is:

`(x - h)^2/a^2 + (y - k)^2/b^2 = 1`

The center is: `(h,k)`
The vertices on the major axis are: `(h - a, k) and (h + a, k)`
The vertices on the minor axis are: `(h, k -b) and (h, k + b)`
The foci are: `(h - sqrt(a^2 - b^2), k) and (h + sqrt(a^2 - b^2), k)`

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

`(x - 3)^2/4^2 + (y - -2)^2/3^2 = 1`

The center is: `(3,-2)`
The vertices on the major axis are: `(-1, -2) and (7, -2)`
The vertices on the minor axis are: `(3, -5) and (3, 1)`

Evaluate: `sqrt(a^2 - b^2) = sqrt(4^2 - 3^2) = sqrt(16 - 9) = sqrt(5)`

The foci are: `(3 - sqrt(5), -2) and (3 + sqrt(5), -2)`

Answer 2

The center is `=(3,-2)`
The vertices are `(7,-2)`, `(-1,-2)`, `(3,1)`, `(3,-5)`
The foci are F`(3-sqrt7,-2)` and F'`(3+sqrt7,-2)`

Explanation:

We compare this equation to

`(x-h)^2/a^2+(y-k)^2/b^2=1`

The center is `(h,k)=(3,-2)`

The vertices are A `(h+a,k)=(7,-2)` and A'`(h-a,k)=(-1,-2)`
and

B `(h,k+b)=(3,1)` and B' `(h,,k-b)=(3,-5)`

To calculate the foci, we need c

`c^2=a^2-b^2=16-9=7`

The foci are F `(h-sqrtc,k)=(3-sqrt7,-2)`
and F' `(h+sqrtc,k)=(3+sqrt7,-2)`

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

Answer 3

Centre is (3, -2), focii are `(-sqrt7 +3, -2) and (sqrt7 +3, -2)`. vertices (on horizontal axis) would be at (-4+3,-2) and (4+3,-2) Or (-1,-2) and (7,-2)

Explanation:

Standard equation of an ellipse centered at (h,k) is `(x-h)^2 / a^2 + (y-k)^2 /b^2 =1` with major axis 2a and minor axis 2b.

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 `c^2= a^2 -b^2`

Comparing the given equation with the standard one, it is seen that a=4, b=3, c=`sqrt(4^2-3^2)= sqrt 7`. Also h= 3, k=-2

Hence Centre is (3, -2), focii are `(-sqrt7 +3, -2) and (sqrt7 +3, -2)`. vertices (on horizontal axis) would be at (-4+3,-2) and (4+3,-2) Or (-1,-2) and (7,-2).