# How do you find the coordinates of the center, foci, the length of the major and minor axis given 16x^2+25y^2+32x-150y=159?

Nov 2, 2016

The center is $= \left(- 1 , 3\right)$
The foci are $\left(2 , 3\right)$ and $\left(- 4 , 3\right)$
The length of the major axis is $= 10$
and the length of the minor axis is $= 8$

#### Explanation:

Rewriting the equation
$16 \left({x}^{2} + 2 x\right) + 25 \left({y}^{2} - 6 y\right) = 159$
$16 \left({x}^{2} + 2 x + 1\right) + 25 \left({y}^{2} - 6 y + 9\right) = 159 + 16 + 225$
$16 {\left(x + 1\right)}^{2} + 25 {\left(y - 3\right)}^{2} = 400$
${\left(x + 1\right)}^{2} / 25 + {\left(y - 3\right)}^{2} / 16 = 1$
Comparing this to the stadard form of the ellipse
${\left(x - {x}_{0}\right)}^{2} / {a}^{2} + {\left(y - {y}_{0}\right)}^{2} / {b}^{2} = 1$
The center is $\left({x}_{0} , {y}_{0}\right) = \left(- 1 , 3\right)$
To calculate the foci, we need $c = \sqrt{{a}^{2} - {b}^{2}}$
$c = \sqrt{25 - 16} = 3$
And the foci are $\left({x}_{0} \pm c , {y}_{0}\right)$
This lead us to F$\left(2 , 3\right)$ and F'$\left(- 4 , 3\right)$
The length of the major axis is $= 10$
and the length of the minor axis is $= 8$

graph{(x+1)^2/25+(y-3)^2/16=1 [-11.625, 10.875, -2.865, 8.385]}