# How do you put 9x^2+25y^2-54x-50y-119=0 in standard form, find the center, the endpoints, vertices, the foci and eccentricity?

Sep 16, 2017

#### Explanation:

${\left(x - h\right)}^{2} / {a}^{2} + {\left(y - k\right)}^{2} / {b}^{2} = 1$ is the equation of an ellipse,

whose center is $\left(h , k\right)$, $2 a$ and $2 b$ are major axis and minor axis (if $a > b$, but if $a < b$, it is reverse). In the former case major axis is parallel to $x$-axis and in latter case major axis is parallel to minor axis.

vertices are $\left(h , k \pm b\right)$ and $\left(h \pm a , k\right)$

Eccentricity $e$ can be obtained using $e = \sqrt{1 - {\left(\text{minor axis")^2/("major axis}\right)}^{2}}$

and focii are at a distance of $a e$ on either side of center along major axis (or $b e$ if major axis is parallel to $y$-axis).

We can write $9 {x}^{2} + 25 {y}^{2} - 54 x - 50 y - 119 = 0$ as

$\left(9 {x}^{2} - 54 x + 81\right) + \left(25 {y}^{2} - 50 y + 25\right) = 119 + 81 + 25 = 225$

or ${\left(3 x - 9\right)}^{2} / 225 + {\left(5 y - 5\right)}^{2} / 225 = 1$

or ${\left(x - 3\right)}^{2} / 25 + {\left(y - 1\right)}^{2} / 9 = 1$ or ${\left(x - 3\right)}^{2} / {5}^{2} + {\left(y + 1\right)}^{2} / {3}^{2} = 1$

center is $\left(3 , 1\right)$, major axis is $2 \times 5 = 10$, which is parallel to $x$-axis and minor axis is $2 \times 3 = 6$

Vertices are $\left(3 , 1 \pm 3\right)$ and $\left(3 \pm 5 , 1\right)$ i.e. $\left(3 , 4\right)$, $\left(3 , - 2\right)$, $\left(8 , 1\right)$ and $\left(- 2 , 1\right)$.

Eccentricity is $e = \sqrt{1 - {\left(\frac{3}{5}\right)}^{2}} = \sqrt{\frac{9}{25}} = \frac{4}{5}$ and $a e = 4$

and focii are $\left(3 \pm 4 , 1\right)$ i.e. $\left(- 1 , 1\right)$and $\left(7 , 1\right)$.

They appear as follows:

graph{(9x^2+25y^2-54x-50y-119)((x-3)^2+(y-1)^2-0.03)((x-3)^2+(y-4)^2-0.03)((x-3)^2+(y+2)^2-0.03)((x-8)^2+(y-1)^2-0.03)((x+2)^2+(y-1)^2-0.03)((x-7)^2+(y-1)^2-0.03)((x+1)^2+(y-1)^2-0.03)=0 [-7.14, 12.86, -3.78, 6.22]}