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
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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) (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) (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:
^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.
Factor out the lead coefficient from each part:
\ +\ 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.
\ +\ 25(y^2\ -\ 8y)\ =\ -144\ +\ 144)
Repeat the process with the 1st degree y-term:
\ +\ 25(y^2\ -\ 8y\ +\ 16)\ =\ -144\ +\ 144\ + 400)
Factor both trinomials:
^2\ +\ 25(y\ -\ 4)^2\ =\ 400)
Divide through by the constant in the RHS
^2}{25}\ +\ \frac{(y\ -\ 4)^2}{16}\ =\ 1)
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
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