Problem source

Show that the point $T$ with coordinates
\[
\left( \frac{a(1-t^2)}{1+t^2} \; , \; \frac{2bt}{1+t^2}\right)
\tag{$*$}
\]
(where $a$ and $b$ are non-zero) lies on the ellipse
\[
\frac{x^2}{a^2} + \frac{y^2}{b^2} =1
\,.
\]
\begin{questionparts}
\item The line $L$ is the tangent to the ellipse at $T$. The point $(X,Y)$ lies on $L$,  and $X^2\ne a^2$. Show that
\[ (a+X)bt^2 -2aYt +b(a-X) =0 \,.\]
Deduce that if $a^2Y^2>(a^2-X^2)b^2$, then there are two distinct lines through $(X,Y)$ that are tangents to the ellipse.  Interpret this result geometrically. Show, by means of a sketch, that the result  holds also  if $X^2=a^2\,$.
\item The distinct points  $P$ and $Q$ are given by $(*)$, with $t=p$ and $t=q$, respectively. The tangents to the ellipse at 
$P$ and $Q$ meet at the point   with coordinates $(X,Y)$, where $X^2\ne a^2\,$.
Show that 
\[ (a+X)pq = a-X\]
 and find an expression for $p+q$ in terms of $a$, $b$, $X$ and $Y$.
Given that the tangents meet the $y$-axis
at points $(0,y_1)$ and $(0,y_2)$, where $y_1+y_2 = 2b\,$, show that
\[
\frac{X^2}{a^2} +\frac{Y}{b}= 1  
\,.
\]
\end{questionparts}

Solution source

\begin{questionparts}
\item The tangent has equation:

\begin{align*}
&& 0 &= \frac{Xx}{a^2} + \frac{Yy}{b^2} -1 \\
\Rightarrow &&&= \frac{Xa(1-t^2)}{a^2(1+t^2)} + \frac{Y2bt}{b^2(1+t^2)} - 1 \\
\Rightarrow &&0&= Xb(1-t^2) + Y2at - ab(1+t^2)\\
&&&= -(b(a+X)t^2 -2aYt +b(a-X)) \\
\Rightarrow && 0 &= (a+X)bt^2-2aYt+b(a-X) \\
\\
&& 0  <\Delta &= 4a^2Y^2 - 4(a+X)b(a-X)b \\
&&&= 4(a^2Y^2-b^2(a^2-X^2)) \\
\Leftrightarrow && a^2Y^2 &> (a^2-X^2)b^2
\end{align*}

Therefore there are two roots to the quadratic, ie two values of the parameter $t$ which works.  The condition is equivalent to $\frac{X^2}{a^2} + \frac{Y^2}{b^2} > 1$. ie from any point outside the ellipse there are two tangent lies.



\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){((#1)+.5)*((#1)-1)*((#1)-2.1)};
    \def\xl{-3};
    \def\xu{3};
    \def\yl{-3};
    \def\yu{3};
    
    % 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);

        % \filldraw (-0.5, 0) circle (1.5pt) node[below]{$p$};
        % \filldraw (1, 0) circle (1.5pt) node[below]{$q$};
        % \filldraw (2.1, 0) circle (1.5pt) node[below]{$r$};

        \def\a{2};
        \def\b{1};

        \filldraw (\a, 0) circle (1.5pt) node[below]{$a$};
        \filldraw (-\a, 0) circle (1.5pt) node[below]{$-a$};
        \filldraw (0, \b) circle (1.5pt) node[right]{$b$};
        \filldraw (0, -\b) circle (1.5pt) node[right]{$-b$};
        
        \draw (\a, \yl) -- (\a, \yu);
        \draw (\xl, {((\a+\xl)*\b*(0.8)^2+\b*(\a-\xl))/2/\a/(0.8)}) -- (\xu,   {((\a+\xu)*\b*(0.8)^2+\b*(\a-\xu))/2/\a/(0.8)}); 
        %  (a+X)bt^2-2aYt+b(a-X) 
        
        \draw[thick, blue, smooth, domain=0:360, samples=100] 
            plot ({\a*cos(\x)}, {\b*sin(\x)});

    \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}

Clearly there are two tangents when $X = \pm a$ (except $(X,Y) = (\pm a, 0)$.

\item We must have $p$ and $q$ are roots of $0 = (a+X)bt^2-2aYt+b(a-X)$, ie $pq = \frac{b(a-X)}{(a+X)b} \Rightarrow (a+X)pq = a-X$. Similarly $p+q = \frac{2aY}{(a+X)b}$

Given that the tangents meet the $y$-axis at $(0, y_i)$ we must have $abt^2-2ay_it + ab = 0$, so

\begin{align*}
&& 0 &= abp^2-2ay_1p + ab \\
&& 0 &= abq^2-2ay_2q + ab \\
\Rightarrow && y_1 &= \frac{ab(p^2+1)}{2ap} \\
&& y_2 &=  \frac{ab(q^2+1)}{2aq} \\
\Rightarrow && 2b &= \frac{ab(p^2+1)}{2ap} +\frac{ab(q^2+1)}{2aq} \\
&&&= \frac{ab(pq(p+q)+p+q)}{2apq} \\
\Rightarrow && 4pq &= pq(p+q)+p+q \\
\Rightarrow && 4 \frac{b(a-X)}{(a+X)b} &= \frac{2aY}{(a+X)b}
 \left ( \frac{b(a-X)}{(a+X)b}  + 1 \right) \\
 &&  &=  \frac{2aY}{(a+X)b}
  \frac{2ab}{(a+X)b} \\
\Rightarrow && 4b^2(a^2-X^2) &= 4a^2bY \\
\Rightarrow && 1 &= \frac{Y}{b} + \frac{X^2}{a^2}
\end{align*}
as required.
\end{questionparts}