1994 Paper 1 Q5

Year: 1994
Paper: 1
Question Number: 5

Course: UFM Pure
Section: Conic sections

Difficulty: 1500.0 Banger: 1516.0

Problem

A parabola has the equation \(y=x^{2}.\) The points \(P\) and \(Q\) with coordinates \((p,p^{2})\) and \((q,q^{2})\) respectively move on the parabola in such a way that \(\angle POQ\) is always a right angle.
  1. Find and sketch the locus of the midpoint \(R\) of the chord \(PQ.\)
  2. Find and sketch the locus of the point \(T\) where the tangents to the parabola at \(P\) and \(Q\) intersect.

Solution

  1. The line \(PO\) has gradient \(\frac{p^2}{p} = p\) and teh line \(QO\) has gradient \(q\), therefore we must have that \(pq = -1\). Therefore, \(R\) is the point \begin{align*} && R &= \left ( \frac{p-\frac{1}{p}}{2}, \frac{p^2+\frac{1}{p^2}}{2} \right) \\ &&&= \left ( \frac12\left ( p - \frac{1}{p} \right),2\left (\frac12 \left(p-\frac{1}{p}\right) \right)^2+1 \right) \\ &&&= \left ( t, 2t^2+1\right) \end{align*} So we are looking at another parabola.
    TikZ diagram
  2. The tangents are \(y = 2px+c\), ie \(p^2 = 2p^2+c\), ie \(y = 2px -p^2\) so we have \begin{align*} && y - 2px &= -p^2 \\ && y - 2qx &= -q^2 \\ \Rightarrow && (2p-2q)x &= p^2-q^2 \\ \Rightarrow && x &= \frac12 (p+q)\\ && y &= p(p+q)-p^2 \\ && y &= pq = -1 \end{align*} Therefore \(x = \frac12(p - \frac1p), y= -1\), so we have the line \(y = -1\) (the directrix)
    TikZ diagram
Rating Information

Difficulty Rating: 1500.0

Difficulty Comparisons: 0

Banger Rating: 1516.0

Banger Comparisons: 1

Show LaTeX source
Problem source
A parabola has the equation $y=x^{2}.$ The points $P$ and $Q$ with coordinates $(p,p^{2})$ and $(q,q^{2})$ respectively move on the parabola in such a way that $\angle POQ$ is always a right angle. 
\begin{questionparts}
\item Find and sketch the locus of the midpoint $R$ of the chord $PQ.$ 
\item Find and sketch the locus of the point $T$ where the tangents to the parabola at $P$ and $Q$ intersect. 
\end{questionparts}
Solution source
\begin{questionparts}
\item  The line $PO$ has gradient $\frac{p^2}{p} = p$ and teh line $QO$ has gradient $q$, therefore we must have that $pq = -1$. Therefore, $R$ is the point
\begin{align*}
&& R &= \left ( \frac{p-\frac{1}{p}}{2}, \frac{p^2+\frac{1}{p^2}}{2} \right) \\
&&&= \left ( \frac12\left ( p - \frac{1}{p} \right),2\left (\frac12 \left(p-\frac{1}{p}\right) \right)^2+1 \right) \\
&&&= \left ( t, 2t^2+1\right)
\end{align*}

So we are looking at another parabola.


\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){1/cos(#1*180/pi)};
    \def\xl{-3};
    \def\xu{3};
    \def\yl{-0.5};
    \def\yu{5.2};
    
    % 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=\xl:\xu, samples=100] 
            plot (\x, {\x*\x});
        \draw[thick, red, smooth, domain=\xl:\xu, samples=100] 
            plot (\x, {2*\x*\x+1});
        \node[below left] at ({pi},0) {$\pi$};

        \draw (1,1) -- (-1,1);
        \filldraw (0,1) circle (1pt);

        \draw(2,4) -- (-0.5, 0.25);
        \filldraw ({(2-0.5)/2},{(4+0.25)*0.5}) circle (1pt);
        
    \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}

\item The tangents are $y = 2px+c$, ie $p^2 = 2p^2+c$, ie $y = 2px -p^2$ so we have

\begin{align*}
&& y - 2px &= -p^2 \\
&& y - 2qx &= -q^2 \\
\Rightarrow && (2p-2q)x &= p^2-q^2 \\
\Rightarrow && x &= \frac12 (p+q)\\
&& y &= p(p+q)-p^2 \\
&& y &= pq = -1
\end{align*}

Therefore $x = \frac12(p - \frac1p), y= -1$, so we have the line $y = -1$ (the directrix)

\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){1/cos(#1*180/pi)};
    \def\xl{-3};
    \def\xu{3};
    \def\yl{-1.5};
    \def\yu{5.2};
    
    % 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=\xl:\xu, samples=100] 
            plot (\x, {\x*\x});
        \draw[thick, red, smooth, domain=\xl:\xu, samples=100] 
            plot (\x, {2*\x*\x+1});

        % \draw (1,1) -- (-1,1);
        \draw (\xl, {2*\xl-1})--(\xu, {2*\xu-1});
        \draw (\xl, {-2*\xl-1})--(\xu, {-2*\xu-1});

        % \draw(2,4) -- (-0.5, 0.25);
        \draw (\xl, {4*\xl-4})--(\xu, {4*\xu-4});
        \draw (\xl, {-1*\xl-0.25})--(\xu, {-1*\xu-0.25});
        % \filldraw ({(2-0.5)/2},{(4+0.25)*0.5}) circle (1pt);

        \draw (\xl,-1) -- (\xu,-1);
        
    \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}

\end{questionparts}