Problem source

Show that the equation $x^3 + px + q=0$ has exactly one real
solution if  $p \ge 0\,$.
A parabola $C$ is given parametrically by
\[
x = at^2, \: \ \ y = 2at \: \: \: \ \ \  \ \ \  \l a > 0 \r \;.
\]
Find an equation which must be satisfied by $t$ at points on $C$ at
which the normal passes through the point $\l h , \; k \r\,$. Hence
show that, if $h \le 2a \,$, exactly one normal to $C$ will pass through
 $\l h , \; k \r \, $.

Find, in Cartesian form, the equation of the locus of the  points
from which exactly two normals can be drawn to $C\,$. Sketch the locus.

Solution source

If $p \geq 0$ then the derivative is $x^2+p \geq 0$ and in particular the function is increasing. Therefore it will have exactly $1$ real root (as for very large negative $x$ it is negative, and vice-versa fo positive $x$).

\begin{align*}
&& \frac{\d y}{\d x} &= \frac{\dot{y}}{\dot{x}} \\
&&&= \frac{2a}{2at} \\
&&&= \frac{1}{t} \\
\text{eq of normal} && \frac{k-2at}{h-at^2} &= -t \\
\Rightarrow && k-2at &= at^3-th \\
&& 0 &= at^3+(2a-h)t-k
\end{align*}

Since $a > 0$ this is the same constraint as the first part, in particular $2a-h \geq 0 \Leftrightarrow 2a \geq h$.

If exactly two normals can be drawn to $C$ we must have that our equation has a repeated root, ie

\begin{align*}
&& 0 &= at^3+(2a-h)t-k\\
&& 0 &= 3at^2+2a-h\\
\Rightarrow && 0 &= 3at^3+ 3(2a-h)t-3k \\
&& 0 &= 3at^3+(2a-h)t \\
\Rightarrow && 0 &= 2(2a-h)t-3k \\
\Rightarrow && t &= \frac{3k}{2(2a-h)} \\
\Rightarrow && 0 &= 3a \left (\frac{3k}{2(2a-h)} \right)^2+2a-h \\
&& 0 &= 27ak^2+4(2a-h)^3
\end{align*}



\begin{center}
    \begin{tikzpicture}
    % \def\functionf(#1){2*(#1)*((#1)^2 - 5)/((#1)^2-4)};
    \def\xl{-3};
    \def\xu{14};
    \def\yl{-10};
    \def\yu{10};
    
    % Calculate scaling factors to make the plot square
    \pgfmathsetmacro{\xrange}{\xu-\xl}
    \pgfmathsetmacro{\yrange}{\yu-\yl}
    \pgfmathsetmacro{\xscale}{10/\xrange}
    \pgfmathsetmacro{\yscale}{10/\yrange}
    
    % Define the styles for the axes and grid
    \tikzset{
        axis/.style={very thick, ->},
        grid/.style={thin, gray!30},
        x=\xscale cm,
        y=\yscale cm
    }
    
    % Define the bounding region with clip
    \begin{scope}
        % You can modify these values to change your plotting region
        \clip (\xl,\yl) rectangle (\xu,\yu);
        
        % Draw a grid (optional)
        % \draw[grid] (-5,-3) grid (5,3);
        
        \draw[thick, blue, smooth, domain=\yl:\yu, samples=201] 
            plot ({2+(27/4*\x*\x)^(1/3)}, {\x});
    \end{scope}

    \filldraw (2,0) circle (1.5pt) node[below] {$2a$};
    
    % Set up axes
    \draw[axis] (\xl,0) -- (\xu,0) node[right] {$x\; (h)$};
    \draw[axis] (0,\yl) -- (0,\yu) node[above] {$y\;(k)$};
    
    \end{tikzpicture}
\end{center}