I need help asap, genuinely don’t understand.

Question

An ellipse or hyperbola uses the general form Ax2+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. x2+Cy2+Dx+Ey+F=0

Given 4 points on a vertical ellipse (-1, -5.473), (0, 0.918), (1, 1.107), and (2.753, 0). Select the missing coefficients (answers have been rounded to the nearest tenth).

x2+     [ Select ]    [".7", ".1", ".3", ".5"]   y2+     [ Select ]    ["-1.3", "-1.0", "-1.7", "-1.9"]  x+     [ Select ]    ["2.7", "2.1", "1.8", "2.5"]  y+     [ Select ]    ["-1.1", "-2.9", "-1.6", "-2.0"]   

Answer 1

Since we are given 4 different coordinates pairs, (x,y), we can plug them in one at a time for x and y into the equation x² + Cy²+ Dx + Ey + F = 0.

Every time we do, we generate a new (linear) equation in the 4 variables, C, D, E, and F.

For example, plugging in the first point, (-1 , -5.473), generates the following equation:

1 + 29.954C - D - 5.473E + F = 0 or 29.954C - 1D - 5.473E + 1F = - 1

Plug in the other 3 points and then solve the resulting linear system in 4 unknowns by whatever method is desirable. Matrices will work easily, but elimination by hand is possible, especially given the zeros.