SOLUTION: Identify the conic. Choose the conic and its center. 16x^2 + 25y^2 - 96x - 200y = -144 I dont understand how to do this problem, ive tried and tried many times but i cant see

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Identify the conic. Choose the conic and its center. 16x^2 + 25y^2 - 96x - 200y = -144 I dont understand how to do this problem, ive tried and tried many times but i cant see      Log On


   



Question 309192: Identify the conic. Choose the conic and its center.
16x^2 + 25y^2 - 96x - 200y = -144
I dont understand how to do this problem, ive tried and tried many times but i cant seem to hack it. Could you please help me step by step to solve this problem? Thank you so very much.

Found 2 solutions by scott8148, solver91311:
Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
the squares of x and y are both positive but the coefficients are different
___ these are the characteristics of an ellipse

rearrange and factor ___ 16(x^2 - 6x) + 25(y^2 - 8y) = -144

complete the squares to find the center ___ 16(x^2 - 6x + 9) + 25(y^2 - 8y + 16) = -144 + 16(9) + 25(16)

16(x - 3)^2 + 25(y - 4)^2 = 400

dividing by 400 ___ {[(x - 3)^2] / 25} + {[(y - 4)^2] / 16} = 1

the ellipse is centered at (3, 4)
___ the semimajor axis is 5 and the semiminor axis is 4

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!


You have both an x-squared term and a y-squared term, therefore it is NOT a parabola.

The signs on the x-squared and the y-squared terms are the same, therefore it is NOT a hyperbola.

The coefficients on the x-squared and y-squared terms are unequal, therefore it is NOT a circle.

Therefore it is an ellipse. Since the coefficient on the x-squared term is smaller than the coefficient on the y-squared term, the ellipse has a horizontal major axis.

In order to determine the center, you need to put the equation into Standard Form. Standard Form for an ellipse with a horizontal major axis and a center at is:



The process is called completing the square.

Arrange the equation so that the x-terms are together and the y-terms are together.



Factor out the lead coefficient from each part:



Take the coefficient of the 1st degree x-term, divide by 2, square the result, add that result inside of the parentheses with the other x-terms, then add that result times the lead coefficient from the x-terms to the RHS. -6 divided by 2 is -3. -3 squared is 9. Add 9 inside of the parentheses with the x-terms, and add 16 times 9 = 144 to the RHS of the equation.



Repeat the process with the 1st degree y-term:



Factor both trinomials:



Divide through by the constant in the RHS




Hence, the center is at While we are at it, recognize that the the semi-major axis measures and the semi-minor axis measures .


John