# How do you find the sum of the coordinates of center in the conic 9x^2+25y^2-18x-150y+9=0?

Dec 19, 2016

Here is a reference Conic section - General Cartesian form that gives the equation:

$A {x}^{2} + B x y + C {y}^{2} + D x + E y + F = 0$

#### Explanation:

Here is a reference Conic section - General Cartesian form that gives the equation:

$A {x}^{2} + B x y + C {y}^{2} + D x + E y + F = 0$

The given equation is:

$9 {x}^{2} + 25 {y}^{2} + 18 x + 150 y + 9 = 0$

We observe that ${B}^{2} - 4 A C = {0}^{2} - 4 \left(9\right) \left(25\right) = - 900$

The reference tells us that this is an ellipse.

The standard Cartesian equation for and ellipse is:

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

where $\left(h , k\right)$ is the center.

Begin the conversion to this form by adding #9h^2 and 25k^2 to both sides of the equation and group all of the x terms and y terms together, respectively:

$9 {x}^{2} - 18 x + 9 {h}^{2} + 25 {y}^{2} - 150 y + 25 {k}^{2} + 9 = 9 {h}^{2} + 25 {k}^{2}$

Remove a common factor of 9 from the first 3 terms and a common factor of 25 from the next 3 terms:

$9 \left({x}^{2} - 2 x + {h}^{2}\right) + 25 \left({y}^{2} - 6 y + {k}^{2}\right) + 9 = 9 {h}^{2} + 25 {k}^{2}$

Using the pattern ${\left(a - b\right)}^{2} = {a}^{2} - 2 a b + {b}^{2}$ we observe that the x terms become a perfect square when $h = 1$ and the y terms become a perfect square when $k = 3$:

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

The 9 on the left and the 9 on the right cancel:

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

Divide both sides by 25(9):

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

Write the denominators as squares:

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

This is an ellipse with the center at $\left(1 , 3\right)$