Question

What is the coordinates of the center of an ellipse defined by the equation 16x^2 + 25y^2 + 160x - 200y + 400 = 0 ? Please give solution .

Answer 1

16x^2 + 25y^2 + 160x - 200y + 400 = 0     Rearrange and regroup.

(16x^2 + 160x) + (25y^2 - 200y ) = 0-400.     Group the xs together and the ys together.

16(X^2 + 10x) + 25(y^2-8y) = -400.     Factorising.

We are going to use completing the square method.

Coefficient of x in the first expression = 10.

Half of it = 1/2 * 10 = 5. (Note this value)

Square it = 5^2  = 25.     (Note this value)


Coefficient of y in the second expression = -8.

Half of it = 1/2 * -8 = -4. (Note this value)

Square it = (-4)^2  = 16. (Note this value)


We are going to carry out a manipulation of completing the square with the values

25 and 16.  By adding and substracting it.


16(X^2 + 10x) + 25(y^2-8y) = -400

16(X^2 + 10x + 25 -25) + 25(y^2-8y + 16 -16) = -400

Note that +25 - 25 = 0.    +16 -16 = 0. So the equation is not altered.

16(X^2 + 10x + 25) -16(25) + 25(y^2-8y + 16) -25(16) = -400


16(X^2 + 10x + 25) + 25(y^2-8y + 16)  = -400 +16(25) + 25(16)    Transferring the terms -16(25) and -25(16)

to other side of equation.  And 16*25 = 400


16(X^2 + 10x + 25) + 25(y^2-8y + 16)  = 25(16)


16(X^2 + 10x + 25) + 25(y^2-8y + 16)  = 400

We now complete the square by using the value when coefficient was halved.


16(x-5)^2 + 25(y-4)^2  = 400

Divide both sides of the equation by 400


(16(x-5)^2)/400 + (25(y-4)^2)/400  = 400/400              Note also that, 16*25 = 400.


((x-5)^2)/25 + ((y-4)^2)/16  = 1

((x-5)^2)/(5^2) + ((y-4)^2)/(4^2)  = 1


Comparing to the general format of an ellipse.

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


Coordinates of the center = (h,k).

Comparing   with above   (x-5) = (x - h) , h = 5.

Comparing   with above   (y-k) = (y - k) , k = 4.

Therefore center = (h,k) = (5,4).

Sorry the answer came a little late.  Cheers.