2.30 \(4xp^{2}=\left ( 3x-b\right ) ^{2}\)

Problem (30)

\[ 4xp^{2}-\left ( 3x-b\right ) ^{2}=0 \]

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

\begin{align*} 0-4\left ( 4x\right ) \left ( -\left ( 3x-b\right ) ^{2}\right ) & =0\\ x\left ( 3x-b\right ) ^{2} & =0 \end{align*}

Hence \(x=0\) (y-axis) is solution. But this does not verify the ode (book is wrong, it said it verifies ode). So can not be envelope. The other solution is \(3x=b\) or \(x=\frac {b}{3}\). Recalling that the \(p\) discriminant has form \(ET^{2}C\) then this y line is Tac locus since it shows twice and \(x=0\) is \(C\) or cusp locus.

The solution to this ode can be found to be

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

Since this is quadratic, then using C-discriminant \(b^{2}-4ac=0\) gives

\begin{align*} \left ( 2y\right ) ^{2}-4\left ( 1\right ) \left ( y^{2}-x\left ( x-b\right ) ^{2}\right ) & =0\\ 4y^{2}-4\left ( y^{2}-x\left ( x-b\right ) ^{2}\right ) & =0\\ y^{2}-y^{2}+x\left ( x-b\right ) ^{2} & =0\\ x\left ( x-b\right ) ^{2} & =0 \end{align*}

Hence \(x=0\) which is same as in \(p\)-discriminant and \(x=b\) which, recalling that C-discriminant is \(EN^{2}C^{3}\) then \(x=b\) is \(N\) or nodal locus.

The following plot shows solution curves (in blue) for different values of \(c_{1}\) with the above solutions. In this plot \(b=6\) we used.

Figure 24: Singular solution envelope with general solution curves