Problem source

Sketch the curve $C_{1}$ whose parametric equations are $x=t^{2},$ $y=t^{3}.$
The circle $C_{2}$ passes through the origin $O$. The points $R$ and $S$ with real non-zero parameters $r$ and $s$ respectively are other intersections of $C_{1}$ and $C_{2}.$ Show that $r$ and $s$ are roots of an equation of the form \[
t^{4}+t^{2}+at+b=0,
\]
where $a$ and $b$ are real constants. 
By obtaining a quadratic equation, with coefficients expressed in terms of $r$ and $s$, whose roots would be the parameters of any further intersections of $C_{1}$ and $C_{2},$ or otherwise, show that $O$, $R$ and $S$ are the only real intersections of $C_{1}$ and $C_{2}.$

Solution source

\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){1/(1+(#1)^2)};
    \def\xl{-5};
    \def\xu{5};
    \def\yl{-5};
    \def\yu{5};
    
    % 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=-5:5, samples=100] 
            plot ({(\x)^2}, {(\x)^3});


    \end{scope}
    
    % Set up axes
    \draw[axis] (\xl,0) -- (\xu,0) node[right] {$x$};
    \draw[axis] (0,\yl) -- (0,\yu) node[above] {$y$};
    
    \end{tikzpicture}
\end{center}

Suppose the circle has centre $(c,d)$, then
\begin{align*}
&& c^2+d^2 &= (t^2-c)^2+(t^3-d)^2 \\
\Rightarrow && 0 &= t^4-2ct^2+t^6-2t^3d \\
\Rightarrow && 0 &= t^4+t^2-2td-2c
\end{align*}

So by setting $a = -2d$ and $b = -2c$ we have the desired equation.

By matching the coefficients of $t^4, t^3, t^2$ we must have:
\begin{align*}
&& 0 &= (t^2-(r+s)t+rs)(t^2+t(r+s)-rs+(r+s)^2+1) \\
\Rightarrow && 0 &= t^2+(r+s)t-rs+(r+s)^2+1 \\
&& \Delta &= (r+s)^2 -4(1-rs+(r+s)^2) \\
&&&= -4+4rs-3(r+s)^2 \\
&&&=-4-2(r+s)^2-(r-s)^2 < 0
\end{align*}

Therefore there are no further (real) solutions. Hence $O, R, S$ are the only solutions.