Mariel M.
asked 11/06/16what are the coefficients?
A horizontal/vertical ellipse or hyperbola use the general form Ax^2+Cy^2+Dx+Ey+F=0Ax2+Cy2+Dx+Ey+F=0. Solving for 5 unknowns (A, B, C, D, E, F) requires 5 equations, needs 5 points given. But if one of the coefficients is divided out (A or C), then only 4 coefficients remain and only 4 points are needed. x^2+Cy^2+Dx+Ey+F=0x2+Cy2+Dx+Ey+F=0
Given 4 points on a horizontal ellipse (3.75, 0), (0, 2.71), (1, -7), and (-1, -5.725). Select the missing coefficients (answers have been rounded to the nearest tenth).
_x2 + _y2 + _x + _y + _ = 0
1 Expert Answer
Raymond B. answered 12d
Math, microeconomics or criminal justice
plug in (3.75=15/4, 0)
225a/16 + 0b + 15c/4 + 0d+e = 0
225a + 60c= -16e
plug in (0, 2.71)
0a + b2.71^2 + 0c + 2.71d + e = 0
7.3441b +2.71d =-e
734.41 +271d = -100e
plug in (1, -7)
a +49b +c -78d= -e
plug in (-1, -5.725=-229/40)
a+b229^2/1600 -c -229d/40 =-e
1600a +52,441b -1600c -9,160d = -1600e
1600a +49x1600b +1600c -78x1600d =-1600e subtract
(52441-78400)b -3200c +(-9160 +124800)d =0 eliminating a and e
-2559b -3200c + 115,640d = 0
continue with substitution elimination and solve for a,b,c,d, and e
or use matrix (linear) algebra with row operations to get a matrix with all 1's on the diagonal and 0's everywhere else, answer then pops up on the last column of the augmented matrix
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Neal D.
11/06/16