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

1 Answer
Dec 19, 2016

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

Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0

Explanation:

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

Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0

The given equation is:

9x^2 + 25y^2 + 18x + 150y + 9 = 0

We observe that B^2 - 4AC = 0^2 - 4(9)(25) = -900

The reference tells us that this is an ellipse.

The standard Cartesian equation for and ellipse is:

(x - h)^2/a^2 + (y - k)^2/b^2 = 1

where (h, k) 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:

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

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

9(x^2 - 2x + h^2) + 25(y^2 - 6y + k^2) + 9 = 9h^2 + 25k^2

Using the pattern (a - b)^2 = a^2 - 2ab + 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(x - 1)^2 + 25(y - 3)^2 + 9 = 9(1)^2 + 25(3)^2

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

9(x - 1)^2 + 25(y - 3)^2 = 25(9)

Divide both sides by 25(9):

(x - 1)^2/25 + (y - 3)^2/9 = 1

Write the denominators as squares:

(x - 1)^2/5^2 + (y - 3)^2/3^2 = 1

This is an ellipse with the center at (1, 3)