SOLUTION: 4x^2 + 9y^2 - 16x + 54y + 96 = 0 a) Find the centre b) the two parameters a and b c) the eccentricity d) the two foci

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: 4x^2 + 9y^2 - 16x + 54y + 96 = 0 a) Find the centre b) the two parameters a and b c) the eccentricity d) the two foci       Log On


   

Question 917854: 4x^2 + 9y^2 - 16x + 54y + 96 = 0
a) Find the centre
b) the two parameters a and b
c) the eccentricity
d) the two foci
Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
4x^2 + 9y^2 - 16x + 54y + 96 = 0
4x^2-16x+9y^2+54y = -96
complete the square:
4(x^2-4x+4)+9(y^2+6y+9) = -96+16+81
4(x-2)^2+9(y+3)^2=1
%28x-2%29%5E2%2F%281%2F4%29%2B%28y%2B3%29%5E2%2F%281%2F9%29=1
This is an equation of an ellipse with horizontal major axis.
Its standard form of equation: %28x-h%29%5E2%2Fa%5E2%2B%28y-k%29%5E2%2Fb%5E2=1, a>b, (h,k)=center
a) Find the centre:(2,-3)
..
b) the two parameters a and b
a^2=1/4
a=1/2
b^2=1/9
b=1/3
..
c) the eccentricity=c/a=.37/.5=0.74
..
d) the two foci
c^2=a^2-b^2=1/4-1/9=9/36-4/36=5/36
c=√5/6≈0.37
foci:(2±c,-3)=(2±.37,-3)=(2.37,-3)and (1.63,-3)