Question 309192
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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 *[tex \Large \left(h, k\right)] is:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{(x\,-\,h)^2}{a^2}\ +\ \frac{(y\,-\,k)^2}{b^2}\ =\ 1]


The process is called completing the square.


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 16x^2\ -\ 96x\ +\ 25y^2\ -\ 200y\ =\ -144]


Factor out the lead coefficient from each part:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 16(x^2\ -\ 6x)\ +\ 25(y^2\ -\ 8y)\ =\ -144]


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.


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 16(x^2\ -\ 6x\ +\ 9)\ +\ 25(y^2\ -\ 8y)\ =\ -144\ +\ 144]


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


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 16(x^2\ -\ 6x\ +\ 9)\ +\ 25(y^2\ -\ 8y\ +\ 16)\ =\ -144\ +\ 144\ + 400]


Factor both trinomials:


*[tex \LARGE \ \ \ \ \ \ \ \ \ \ 16(x\ -\ 3)^2\ +\ 25(y\ -\ 4)^2\ =\ 400]


Divide through by the constant in the RHS



*[tex \LARGE \ \ \ \ \ \ \ \ \ \ \frac{(x\ -\ 3)^2}{25}\ +\ \frac{(y\ -\ 4)^2}{16}\ =\ 1]


Hence, the center is at *[tex \Large \left(3, 4\right)]  While we are at it, recognize that the the semi-major axis measures *[tex \Large \sqrt{25}\ =\ 5] and the semi-minor axis measures *[tex \Large \sqrt{16}\ =\ 4].



John
*[tex \LARGE e^{i\pi} + 1 = 0]
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