SOLUTION: 9(x+1)^2+25(y-2)^2=225 what is center: Foci: legnth of major axis Legnth of minor axis standard form Also same for 8x^2+2y^2=32

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: 9(x+1)^2+25(y-2)^2=225 what is center: Foci: legnth of major axis Legnth of minor axis standard form Also same for 8x^2+2y^2=32      Log On


   

Question 451202: 9(x+1)^2+25(y-2)^2=225
what is
center:
Foci:
legnth of major axis
Legnth of minor axis
standard form
Also same for 8x^2+2y^2=32
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi,

9(x+1)^2+25(y-2)^2=225

%28x%2B1%29%5E2%2F25+%2B%28y-2%29%5E2%2F9+=+1

 Standard Form of an Equation of an Ellipse is %28x-h%29%5E2%2Fa%5E2+%2B+%28y-k%29%5E2%2Fb%5E2+=+1+

where Pt(h,k) is the center and a and b  are the respective vertices distances from center.

center: (-1,2)

Foci: sqrt%2825-9%29 = 4  Foci (-4,2) and ( 2,2)

length of major axis = 10  (a = 5)  

Length of minor axis = 6   (b = 3)





8x^2+2y^2=32

x%5E2%2F4+%2B+y%5E2%2F16+=+1 C(0,0) major axis = 8 and minor = 4

foci are (0, ± sqrt(12))  sqrt(16-4)