# SOLUTION: Given the ellipse: 2x^2 + 6y^2 + 32x - 48y + 212 = 0 find. a. The Center C b. Length of Major Axis c. Length of Minor Axis d. Distance from C to foci

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 Algebra: Conic sections - ellipse, parabola, hyperbola Solvers Lessons Answers archive Quiz In Depth

 Question 259713: Given the ellipse: 2x^2 + 6y^2 + 32x - 48y + 212 = 0 find. a. The Center C b. Length of Major Axis c. Length of Minor Axis d. Distance from C to fociAnswer by stanbon(57967)   (Show Source): You can put this solution on YOUR website!Given the ellipse: 2x^2 + 6y^2 + 32x - 48y + 212 = 0 Complete the square on the x-terms and on the y-terms: 2(x^2+16x+?) + 6(y^2-8y+?) = -212+? 2(x^2+16x+64) + 6(y^2-8y+16) = -212+2*64+6*16 2(x+8)^2 + 6(y-4) = 12 Divide thru by 12 to get: (x+8)^2/6 + (y-4)^2/2 = 1 ------ h = -8 ; k = 4 a = sqrt(6) ; b = sqrt(2) ------------------------ find. a. The Center C ..........(h,k) b. Length of Major Axis...2a c. Length of Minor Axis,,,2b d. Distance from C to foci: a^2 + b^2 = distance^2 d = sqrt(6 + 2) d = 2sqrt(2) ======================================== Cheers, Stan H. ------------------------------