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

Answer:

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)#