Problem source

The point $P(a\cos\theta\,,\, b\sin\theta)$, where $a>b>0$, lies on the ellipse 
\[\dfrac {x^2}{a^2} + \dfrac {y^2}{b^2}=1\,.\]
The point $S(-ea\,,\,0)$, where $b^2=a^2(1-e^2)\,$, is a focus of the ellipse. The point $N$ is the foot of the perpendicular from  the origin, $O$, to the tangent to the ellipse at $P$. The lines $SP$ and $ON$ intersect at $T$. Show that the $y$-coordinate of  $T$ is 
\[\dfrac{b\sin\theta}{1+e\cos\theta}\,.\]
Show that $T$ lies on the circle with centre $S$ and radius $a$.

Solution source

Find the gradient of the tangent of the ellipse at $P$:

\begin{align*}
&& \frac{2x}{a^2} + \frac{2y}{b^2} \frac{\d y}{\d x} &= 0 \\
\Rightarrow && \frac{\d y}{\d x} &= - \frac{2xb^2}{2ya^2} \\
&&&=- \frac{a \cos \theta b^2}{b \sin \theta a^2} \\
&&&=-\frac{b}{a} \cot \theta
\end{align*}

Therefore the gradient of $ON$ is $\frac{a}{b} \tan \theta$.

\begin{align*}
&& y &= \frac{a}{b} \tan \theta x \\
&& \frac{y-0}{x-(-ea)} &= \frac{b\sin \theta-0}{a\cos \theta -(-ea)} \\
&& y &= \frac{b \sin \theta}{a(e+\cos \theta)}(x+ea) \\
\Rightarrow && y &= \frac{b \sin \theta}{a(\cos \theta+e)}\frac{b}{a} \cot \theta y+ \frac{eb \sin \theta}{\cos \theta + e} \\
&&&= \frac{b^2 \cos \theta}{a^2(\cos \theta +e)}y + \frac{eb \sin \theta}{\cos \theta + e} \\
\Rightarrow && (\cos \theta+e)y &= (1-e^2)\cos \theta y +eb \sin \theta\\
&& e(1+e\cos \theta)y &= eb \sin \theta \\
\Rightarrow && y &= \frac{b \sin \theta}{1+e\cos \theta} \\
&& x &= \frac{b \sin \theta}{1+e\cos \theta} \frac{b}{a} \cot \theta \\
&&&= \frac{b^2 \cos \theta}{a(1+e\cos \theta)}
\end{align*}

Therefore $\displaystyle T\left (\frac{b^2 \cos \theta}{a(1+e\cos \theta)}, \frac{b \sin \theta}{1+e\cos \theta} \right)$.

Finally, we can look at the distance of $T$ from $S$

\begin{align*}
&& d^2 &= \left (\frac{b^2 \cos \theta}{a(1+e\cos \theta)}-(-ea) \right)^2 + \left (\frac{b \sin \theta}{1+e\cos \theta} -0\right)^2 \\
&&&= \frac{\left (b^2 \cos \theta+ea^2(1+e\cos\theta)\right)^2 + \left ( ab \sin \theta\right)^2}{a^2(1+e\cos \theta)^2} \\
&&&= \frac{b^4\cos^2\theta+e^2a^4(1+e\cos\theta)^2+2ea^2b^2(1+e\cos\theta)+a^2b^2\sin^2\theta}{a^2(1+e\cos\theta)^2} \\
&&&= \frac{a^4(1-e^2)^2\cos^2\theta+e^2a^4(1+e\cos\theta)^2+2ea^2a^2(1-e^2)(1+e\cos\theta)+a^4(1-e^2)\sin^2\theta}{a^2(1+e\cos\theta)^2} \\
&&&= a^2 \left ( \frac{(1-e^2)^2\cos^2\theta+e^2(1+e\cos\theta)^2+2e(1-e^2)(1+e\cos\theta)+(1-e^2)(1-\cos^2\theta)}{(1+e\cos\theta)^2} \right) \\
&&&=  a^2 \left ( \frac{e^2(1+e\cos\theta)^2+(1-e^2)((1-e^2)\cos^2\theta+2e(1+e\cos\theta)+(1-\cos^2\theta))}{(1+e\cos\theta)^2} \right) \\
&&&=  a^2 \left ( \frac{e^2(1+e\cos\theta)^2+(1-e^2)(1+e\cos\theta)^2}{(1+e\cos\theta)^2} \right) \\
&&&= a^2
\end{align*}

Therefore a circle radius $a$ centre $S$.

\begin{center}
    \begin{tikzpicture}
        \def\xl{-4};
        \def\xu{4};
    
        \draw[->] (-4,0) -- (4,0);
        \draw[->] (0,-4) -- (0,4);

        \def\a{3.5};
        \def\b{2.2};
        \def\e{sqrt(1-\b*\b/(\a*\a))};

        \draw[thick, blue, smooth, domain=0:360, samples=100] 
            plot ({\a*cos(\x)},{\b*sin(\x)});

        \filldraw ({-\e*\a},0) circle (1pt) node[below] {$S(-ea,0)$};

        \draw ({-\e*\a},0) circle (\a);

        \begin{scope}
        % You can modify these values to change your plotting region
            \clip (\xl,\xl) rectangle (\xu,\xu);
            \draw (\xl, {\xl*\a/\b*tan(45)}) -- (\xu, {\xu*\a/\b*tan(45)});
            \draw (\xl, {(\b*sin(45)/(\a*(cos(45)+\e))*(\xl+\e*\a)}) -- (\xu, {(\b*sin(45)/(\a*(cos(45)+\e))*(\xu+\e*\a)});
        \end{scope}
    \end{tikzpicture}
\end{center}