what are the coefficients?

Question

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

Answer 1

Ax²+Cy²+Dx+Ey+F=0 is indeed the general equation. And you have been given 4 points, so assume A = 1. Next, substitute the x and y values from the first point, giving an equation involving only (at worst) unknowns B, C, D & E. Repeat this with each of the three other given ordered pairs.

 

In the end, you have system of 4 linear equations and those four capital letters as the unknowns. Personally, I think you could use the "rref" method on a TI-84 calculator, operating on the augmented system matrix as the means of solving, but it could be that you'd prefer a tedious non-technology method.