Question 64723: find the foci of the ellipse: 16x^2 + 25y^2 - 96x + 200y + 144 =0
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! find the foci of the ellipse: 16x^2 + 25y^2 - 96x + 200y + 144 =0
16(x^2-(96/16)x+?)+25(y^2+(200/25)y+?)=-144
16(x^2-6x+3^2) + 25(y^+8y+4^2)=-144+16*3^2+25(4^2)
16(x+(3/16))^2+25(y+4)^2=400
[(x+(3/16))^2]/[400/16]+[(y+4)^2]/[400/25]=1
a=[20/4]=5 ; b=[20/5]=4
For an ellipse a^2=b^2+c^2
25=16+c^2
c=3
Center of this ellipse is at (-3/16,-4)
Foci are at ((-3/16)+3,-4) and ((-3/16)-3,-4)
Cheers,
Stan H.
|
|
|
