I've tried 2 hours to do this so I hope someone can help me:
$$11400000=-0.64x^2+2560x-y^2+6000y$$
It says that it have to equal an ellipse with center at the point $(2000,3000)$ and a horizontal semi-axis $a=500$ and a vertical semi-axis $b=400$
To prove that my teacher say I have turn the general form into the standard form.
I'm able to get the center at $(2000, 3000)$ but I don t get $a= 500$ and $b=400$
Thank you in advance!
You don't need quadratic programming, just completing the square would suffice
$ -0.64x^2 + 2560x = -0.64(x^2 - 4000x) = -0.64(x^2 - 4000x + 2000^2 - 2000^2) = -0.64(x^2 - 4000x + 2000^2) + 0.64 \times 2000^2 = -0.64(x - 2000)^2 + 2560000 $
$ - y^2 + 6000y = -1(y^2 - 6000y) = -1(y^2 - 6000y + 3000^2 - 3000^2) = -1(y^2 - 6000y + 3000^2) + 3000^2 = -(y - 3000)^2 + 9000000 $
Therefore we have
$ 11400000=-0.64x^2+2560x-y^2+6000y $
$ 11400000 = (-0.64(x - 2000)^2 + 2560000) + (-(y - 3000)^2 + 9000000) $
$ 0.64(x - 2000)^2 + (y - 3000)^2 = 160000 $
$ \frac{0.64(x - 2000)^2}{160000} + \frac{(y - 3000)^2}{160000} = 1 $
$ \frac{(x - 2000)^2}{500^2} + \frac{(y - 3000)^2}{400^2} = 1 $