How do you find the center, foci and vertices of 9x^2+25y^2-36x-50y+61=0?

Jan 14, 2017

The center is C $= \left(2 , 1\right)$
The foci are F$= \left(\frac{10}{3} , 1\right)$and F'$= \left(\frac{2}{3} , 1\right)$
The vertices are A$= \left(\frac{11}{3} , 1\right)$ and A'$= \left(\frac{1}{3} , 1\right)$ and B$= \left(2 , 2\right)$ and B'$= \left(2 , 0\right)$

Explanation:

Let 's rearrange the equation by completing the squares

$9 {x}^{2} + 25 {y}^{2} - 36 x - 50 y + 61 = 0$

$9 {x}^{2} - 36 x + 25 {y}^{2} - 50 y = - 61$

$9 \left({x}^{2} - 4 x\right) + 25 \left({y}^{2} - 2 y\right) = - 61$

$9 \left({x}^{2} - 4 x + \textcolor{red}{4}\right) + 25 \left({y}^{2} - 2 y + \textcolor{b l u e}{1}\right) = - 61 + \textcolor{red}{36} + \textcolor{b l u e}{50}$

$9 {\left(x - 2\right)}^{2} + 25 {\left(y - 1\right)}^{2} = 25$

Dividing throughout by $25$

$9 {\left(x - 2\right)}^{2} / 25 + 25 {\left(y - 1\right)}^{2} / 25 = \frac{25}{25}$

${\left(x - 2\right)}^{2} / {\left(\frac{5}{3}\right)}^{2} + {\left(y - 1\right)}^{2} / 1 = 1$

This is the equation of an ellipse

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

The center is C $\left(h , k\right) = \left(2 , 1\right)$

$c = \sqrt{{a}^{2} - {b}^{2}} = \sqrt{\frac{25}{9} - 1} = \sqrt{\frac{16}{9}} = \frac{4}{3}$

The foci are F$\left(h + c , k\right) = \left(2 + \frac{4}{3} , 1\right) = \left(\frac{10}{3} , 1\right)$

and F'$\left(h - c , k\right) = \left(2 - \frac{4}{3} , 1\right) = \left(\frac{2}{3} , 1\right)$

graph{(x-2)^2/(25/9)+(y-1)^2=1 [-5.546, 5.55, -2.773, 2.774]}