# How do you find the center, foci and vertices of 16x^2+25y^2-32x+50y+16=0?

Dec 18, 2017

Center is $\left(1 , - 1\right)$, focii are $\left(1 \pm \frac{3}{4} , - 1\right)$ i.e. $\left(\frac{1}{4} , - 1\right)$ and $\left(\frac{7}{4} , - 1\right)$ and vertices are $\left(1 , 0\right)$, $\left(1 , - 2\right)$, $\left(- \frac{1}{4} , - 1\right)$ and $\left(\frac{9}{4} , - 1\right)$

#### Explanation:

As coefficients of ${x}^{2}$ and ${y}^{2}$ are positive but different, it is an ellipse and to find center, focii and vertices, let us convert it into the form ${\left(x - h\right)}^{2} / {a}^{2} + {\left(y - k\right)}^{2} / {b}^{2} = 1$, where $\left(h , k\right)$ is center and major and minor axis are $2 a$ and $2 b$, (the one which is greater is major axis and smaller one is minor axis).

Vertices are $\left(h , k \pm b\right)$ and $\left(h \pm a , k\right)$ and focii are given by $\left(h \pm a e , k\right)$ if $a > b$ and $\left(h , k \pm b e\right)$ if $a < b$, where $e$ is eccentricity given by $\sqrt{1 - {b}^{2} / {a}^{2}}$ or $\sqrt{1 - {a}^{2} / {b}^{2}}$.

$16 {x}^{2} + 25 {y}^{2} - 32 x + 50 y + 16 = 0$

or $16 {x}^{2} - 32 x + 25 {y}^{2} + 50 y + 16 = 0$

or $16 \left({x}^{2} - 2 x + 1 - 1\right) + 25 \left({y}^{2} + 2 y + 1 - 1\right) + 16 = 0$

or $16 {\left(x - 1\right)}^{2} + 25 {\left(y + 1\right)}^{2} - 16 - 25 + 16 = 0$

or $16 {\left(x - 1\right)}^{2} + 25 {\left(y + 1\right)}^{2} = 25$

or ${\left(x - 1\right)}^{2} / \left(\frac{25}{16}\right) + {\left(y + 1\right)}^{2} / 1 = 1$

or ${\left(x - 1\right)}^{2} / {\left(\frac{5}{4}\right)}^{2} + {\left(y + 1\right)}^{2} / {1}^{2} = 1$

Hence, center is $\left(1 , - 1\right)$ and as $a > b$, major axis is $\frac{5}{2}$ and minor axis is $2$.

Vertices are $\left(1 , - 1 \pm 1\right)$ and $\left(1 \pm \frac{5}{4} , - 1\right)$ i.e. $\left(1 , 0\right)$, $\left(1 , - 2\right)$, $\left(- \frac{1}{4} , - 1\right)$ and $\left(\frac{9}{4} , - 1\right)$

Eccentricity is $\sqrt{1 - {1}^{2} / {\left(\frac{5}{4}\right)}^{2}} = \sqrt{1 - \frac{16}{25}} = \frac{3}{5} = 0.6$

and $a e = \frac{5}{4} \times \frac{3}{5} = \frac{3}{4}$ and hence focii are $\left(1 \pm \frac{3}{4} , - 1\right)$ i.e. $\left(\frac{1}{4} , - 1\right)$ and $\left(\frac{7}{4} , - 1\right)$

graph{(16x^2+25y^2-32x+50y+16)((x-0.25)^2+(y+1)^2-0.002)((x-1.75)^2+(y+1)^2-0.002)((x-1)^2+y^2-0.002)((x-1)^2+(y+2)^2-0.002)((x+0.25)^2+(y+1)^2-0.002)((x-2.25)^2+(y+1)^2-0.002)((x-1)^2+(y+1)^2-0.002)=0 [-1.48, 3.52, -2.22, 0.28]}