\blue{x_P} =
\blue{y_P} =
Consider the function f: \mathbb R^2 \to \mathbb R with
f(x,y) = A x^3 + Bx y+ Cx + D y^2.
This has two critical points
P = (\blue{x_P} ,\blue{y_P}) and
Q = (\red{x_Q} ,\red{y_Q}). Find the coordinates.
\blue{x_P} =
\blue{y_P} =
\red{x_Q} =
\red{y_Q}=
A critical point (x_0,y_0) satisfies f_x (x_0,y_0) = 0 = f_y (x_0,y_0).
We compute also I: \ f_x(x,y)= 3*A x^2 + B y + C and
II:\ f_y(x,y)= B x+ fractionReduce(2* B*B, 6*A*(L1+L2)) y.
Therfore we are seeking for pairs x,y such that I and II are zero.
Using II we get the condition y = fractionReduce( -3*A*(L1+L2),B ) x.
Now we substitute this in I.
This transforms I into a quadratic equation:
3*A x^2 + -3*A*(L1+L2) y + C = 0, die sich vereinfacht
zu
x^2 + -(L1+L2) y + L1*L2 = 0.
By factorisation
x^2 + -(L1+L2) y + L1*L2 = (x - L1) (x - L2)
we read the x - coordinates x_P = L1 and
x_Q = L2.
Putting these into the equation II above
y = fractionReduce( -3*A*(L1+L2),B ) x yields the y - coordinates
y_P = Y1 and
y_Q = Y2.