2.31 \(4p^{2}x\left ( x-A\right ) \left ( x-B\right ) =\left ( 3x^{2}-2x\left ( A+B\right ) +AB\right ) ^{2}\)

Problem (31)

\[ 4p^{2}x\left ( x-A\right ) \left ( x-B\right ) -\left ( 3x^{2}-2x\left ( A+B\right ) +AB\right ) ^{2}=0 \]

The ode is quadratic in \(p\). Using the quadratic discriminant \(b^{2}-4ac=0\) gives

\begin{align*} 0-4\left ( 4x\left ( x-A\right ) \left ( x-B\right ) \right ) \left ( -\left ( 3x^{2}-2x\left ( A+B\right ) +AB\right ) ^{2}\right ) & =0\\ -16x\left ( x-A\right ) \left ( x-B\right ) \left ( 3x^{2}-2x\left ( A+B\right ) +AB\right ) ^{2} & =0\\ x\left ( x-A\right ) \left ( x-B\right ) \left ( 3x^{2}-2x\left ( A+B\right ) +AB\right ) ^{2} & =0 \end{align*}

Hence lines are \(x=0,x=A,x=B\). The ode is not satisfied by these (again, book says they are. Book must be wrong). So looking at \(ET^{2}C\), then these are Cusp locus. Looking at the second factor \(\left ( 3x^{2}-2x\left ( A+B\right ) +AB\right ) ^{2}=0\), this gives \(\left ( 3x^{2}-2x\left ( A+B\right ) +AB\right ) =0\) (factor is 2), or

\[ 3x^{2}-2x\left ( A+B\right ) +AB=0 \]

The solution is

\begin{equation} x=\frac {2\left ( A+B\right ) \pm \sqrt {4\left ( A+B\right ) ^{2}-12AB}}{6}\tag {1}\end{equation}

Since it shows as factor of two, and comparing to \(ET^{2}C\,\,\) shows the above is Tac locus.  The solution to the ode can be found to be

\[ \left ( y+c_{1}\right ) ^{2}=x\left ( x-A\right ) \left ( x-B\right ) \]

The following plot is made using \(A=3,B=6\). It shows \(x=0,x=A,x=B\) as lines parallel to the y axis. These are Cusp locus (book says these are Envelope, but these do not satisfy the ode. So something is wrong in book, or I am overlooking something). There are two additional vertical lines that comes from Eq (1). these are Tac locus. One of these two lines is real contact line. The other the book calls imaginary point of contact. But both are drawn below.

Figure 25: Singular solution envelope with general solution curves