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

Nov 22, 2016

Please see the explanation.

#### Explanation:

The standard form for the equation of an ellipse is:

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

The center is: $\left(h , k\right)$
The vertices on the major axis are: $\left(h - a , k\right) \mathmr{and} \left(h + a , k\right)$
The vertices on the minor axis are: $\left(h , k - b\right) \mathmr{and} \left(h , k + b\right)$
The foci are: $\left(h - \sqrt{{a}^{2} - {b}^{2}} , k\right) \mathmr{and} \left(h + \sqrt{{a}^{2} - {b}^{2}} , k\right)$

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

${\left(x - 3\right)}^{2} / {4}^{2} + {\left(y - - 2\right)}^{2} / {3}^{2} = 1$

The center is: $\left(3 , - 2\right)$
The vertices on the major axis are: $\left(- 1 , - 2\right) \mathmr{and} \left(7 , - 2\right)$
The vertices on the minor axis are: $\left(3 , - 5\right) \mathmr{and} \left(3 , 1\right)$

Evaluate: $\sqrt{{a}^{2} - {b}^{2}} = \sqrt{{4}^{2} - {3}^{2}} = \sqrt{16 - 9} = \sqrt{5}$

The foci are: $\left(3 - \sqrt{5} , - 2\right) \mathmr{and} \left(3 + \sqrt{5} , - 2\right)$

Nov 22, 2016

The center is $= \left(3 , - 2\right)$
The vertices are $\left(7 , - 2\right)$, $\left(- 1 , - 2\right)$, $\left(3 , 1\right)$, $\left(3 , - 5\right)$
The foci are F$\left(3 - \sqrt{7} , - 2\right)$ and F'$\left(3 + \sqrt{7} , - 2\right)$

#### Explanation:

We compare this equation to

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

The center is $\left(h , k\right) = \left(3 , - 2\right)$

The vertices are A $\left(h + a , k\right) = \left(7 , - 2\right)$ and A'$\left(h - a , k\right) = \left(- 1 , - 2\right)$
and

B $\left(h , k + b\right) = \left(3 , 1\right)$ and B' $\left(h , , k - b\right) = \left(3 , - 5\right)$

To calculate the foci, we need c

${c}^{2} = {a}^{2} - {b}^{2} = 16 - 9 = 7$

The foci are F $\left(h - \sqrt{c} , k\right) = \left(3 - \sqrt{7} , - 2\right)$
and F' $\left(h + \sqrt{c} , k\right) = \left(3 + \sqrt{7} , - 2\right)$

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

Nov 22, 2016

Centre is (3, -2), focii are $\left(- \sqrt{7} + 3 , - 2\right) \mathmr{and} \left(\sqrt{7} + 3 , - 2\right)$. 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 ${\left(x - h\right)}^{2} / {a}^{2} + {\left(y - k\right)}^{2} / {b}^{2} = 1$ 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 $\left(- \sqrt{7} + 3 , - 2\right) \mathmr{and} \left(\sqrt{7} + 3 , - 2\right)$. vertices (on horizontal axis) would be at (-4+3,-2) and (4+3,-2) Or (-1,-2) and (7,-2).