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A parallelogram circumscribes the ellipse and two of its opposite angular points lie on the straight lines ${{x}^{2}}={{h}^{2}}$ ; prove that the locus of the other two is the conic
$\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}\left( 1-\dfrac{{{a}^{2}}}{{{h}^{2}}} \right)=1$.

Answer Verified Verified
Hint: Find vertices of parallelogram by intersection of tangents (sides of parallelogram) using parametric coordinates of point of contacts.

We have ellipse given;
$\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1................\left( i \right)$
As, given in question, two of the angular points of parallelogram is lying on ${{x}^{2}}={{h}^{2}}$.


$ABCD$is a parallelogram.
Let $AB,BC,CD,DA$ is touching ellipse at points $P,Q,R,S$ respectively. In other language, we can say that two tangents are drawn from points $P\text{ and }S,$ two tangents are from $P\text{ and }Q,$ two from $Q\text{ and }R;$ similarly, two tangents from $R\text{ and }S,$ which are intersecting at $A,B,C\And D$ respectively.
Let two points which are lying on ${{x}^{2}}={{h}^{2}}$are $B$and $D$.
Let point $P\text{ and }Q$ is $\left( a\cos \alpha ,b\sin \alpha \right)$ and $\left( a\cos \beta ,b\sin \beta \right)$ as parametric coordinates on ellipse $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1.$
We know that tangent from any point $\left( {{x}_{1}},{{y}_{1}} \right)$ on ellipse is;
$T=0$
Or $\dfrac{x{{x}_{1}}}{{{a}_{2}}}+\dfrac{y{{y}_{1}}}{{{b}_{2}}}=1$
Hence, tangents through $P\left( a\cos \alpha ,b\sin \alpha \right)$ and $Q\left( a\cos \beta ,b\sin \beta \right)$;
$\dfrac{x\cos \alpha }{a}+\dfrac{y\sin \alpha }{b}=1\left( PB \right).............\left( ii \right)$
$\dfrac{x\cos \beta }{a}+\dfrac{y\sin \beta }{b}=1\left( BQ \right).............\left( iii \right)$
Let us find intersecting point of equation (ii) and (iii);
By cross multiplication method from equation (ii) and (iii);
$\begin{align}
  & \dfrac{x}{\dfrac{\sin \alpha }{b}-\dfrac{\sin \beta }{b}}=\dfrac{-y}{\dfrac{\cos \alpha }{a}-\dfrac{\cos \beta }{a}}=\dfrac{-1}{\dfrac{\cos \alpha }{a}\dfrac{\sin \beta }{b}-\dfrac{\sin \alpha \cos \beta }{ab}} \\
 & \dfrac{bx}{\left( \sin \alpha -\sin \beta \right)}=\dfrac{-ay}{\cos \alpha -\cos \beta }=\dfrac{-ab}{\cos \alpha \sin \beta -\sin \alpha \cos \beta } \\
\end{align}$
Simplifying the above relation to get $x$ and $y$
$x=\dfrac{a\left( \sin \alpha -\sin \beta \right)}{\sin \alpha \cos \beta -\cos \alpha \sin \beta },y=\dfrac{-b\left( \cos \alpha -\cos \beta \right)}{\sin \alpha \cos \beta -\cos \alpha \sin \beta }$
We have;
$\begin{align}
  & \sin C-\sin D=2\sin \dfrac{C-D}{2}\cos \dfrac{C+D}{2} \\
 & \cos C-\cos D=-2\sin \dfrac{C-D}{2}\sin \dfrac{C+D}{2} \\
 & \sin C\cos D-\cos C\sin D=\sin \left( C-D \right) \\
\end{align}$
Therefore, we can rewrite $x$ and $y$as;
$x=\dfrac{2a\sin \left( \dfrac{\alpha -\beta }{2} \right)\cos \left( \dfrac{\alpha +\beta }{2} \right)}{\sin \left( \alpha -\beta \right)},y=\dfrac{2b\sin \dfrac{\alpha -\beta }{2}\sin \dfrac{\alpha +\beta }{2}}{-\sin \left( \alpha -\beta \right)}$
We have;
\[\begin{align}
  & \sin 2\theta =2\sin \theta \cos \theta \\
 & x=\dfrac{2a\sin \left( \dfrac{\alpha -\beta }{2} \right)\cos \left( \dfrac{\alpha +\beta }{2} \right)}{2\sin \left( \dfrac{\alpha -\beta }{2} \right)\cos \left( \dfrac{\alpha -\beta }{2} \right)}=\dfrac{a\cos \left( \dfrac{\alpha +\beta }{2} \right)}{\cos \left( \dfrac{\alpha -\beta }{2} \right)} \\
 & y=\dfrac{2b\sin \left( \dfrac{\alpha -\beta }{2} \right)\sin \left( \dfrac{\alpha +\beta }{2} \right)}{2\sin \left( \dfrac{\alpha -\beta }{2} \right)\cos \left( \dfrac{\alpha -\beta }{2} \right)}=\dfrac{b\sin \left( \dfrac{\alpha +\beta }{2} \right)}{\cos \left( \dfrac{\alpha -\beta }{2} \right)} \\
 & \left( x,y \right)=\left( \dfrac{a\cos \left( \dfrac{\alpha +\beta }{2} \right)}{\cos \left( \dfrac{\alpha -\beta }{2} \right)},\dfrac{b\sin \left( \dfrac{\alpha +\beta }{2} \right)}{\cos \left( \dfrac{\alpha -\beta }{2} \right)} \right)..........\left( iv \right) \\
\end{align}\]
Calculated $\left( x,y \right)$ in equation (iv) will lie on ${{x}^{2}}={{h}^{2}}$which is intersection of $PB$ and $BQ$ .
Let us find out points $A$ or $C$ of which we need to find locus.
Let us find out the intersection of $QC$ and $RC$ which is point $C$. Let's coordinate the point $C$ is $\left( {{x}_{_{1}}},{{y}_{1}} \right)$.
The parametric coordinate of point $Q$ is $\left( a\cos \beta ,b\sin \beta \right)$.
Now, by symmetry point $P$ and $R$ will lie exactly opposite to other as shown below;

Hence, parametric coordinate of point $R$is $a\cos \left( \pi +\alpha \right),b\sin \left( \pi +\alpha \right)\text{ or }-a\cos \alpha ,-b\sin \alpha .$
Now, let us write equation of tangents passing through $C$ i.e., $RC$ and $QC$ ;
$\begin{align}
  & T=0 \\
 & \dfrac{-xa\cos \alpha }{{{a}^{2}}}\dfrac{-yb\sin \alpha }{{{b}^{2}}}=1 \\
 & \dfrac{x\cos \alpha }{a}+\dfrac{y\sin \alpha }{b}+1=0.............\left( v \right) \\
\end{align}$
Another tangent through $Q$ is;
$\dfrac{x\cos \beta }{a}+\dfrac{y\sin \beta }{b}-1=0.............\left( vi \right)$
Let us find out intersection of above two equations by cross multiplication method:
$\begin{align}
  & \dfrac{x}{\dfrac{-\sin \alpha }{b}-\dfrac{\sin \beta }{b}}=\dfrac{-y}{\dfrac{-\cos \alpha }{a}-\dfrac{\cos \beta }{a}}=\dfrac{1}{\dfrac{\cos \alpha }{a}\dfrac{\sin \beta }{b}-\dfrac{\sin \alpha }{b}\dfrac{\cos \beta }{a}} \\
 & \dfrac{-bx}{\sin \alpha +\sin \beta }=\dfrac{ay}{\cos \alpha +\cos \beta }=\dfrac{ab}{\cos \alpha \sin \beta -\sin \alpha \cos \beta } \\
\end{align}$
Now, we can write values of $x$ and $y$ as;
$\begin{align}
  & x=\dfrac{a\left( \sin \alpha +\cos \beta \right)}{\sin \alpha \cos \beta -\cos \alpha \sin \beta } \\
 & y=\dfrac{b\left( \cos \alpha +\cos \beta \right)}{\cos \alpha \sin \beta -\sin \alpha \cos \beta } \\
\end{align}$
We have,
$\begin{align}
  & \sin C+\sin D=2\sin \dfrac{C+D}{2}\cos \dfrac{C-D}{2} \\
 & \cos C+\cos D=-2\cos \dfrac{C+D}{2}\cos \dfrac{C-D}{2} \\
 & \sin C\cos D-\cos C\sin D=\sin \left( C-D \right) \\
\end{align}$
Therefore, we can write $x$ and $y$ as;
$x=\dfrac{2a\sin \dfrac{\alpha +\beta }{2}\cos \dfrac{\alpha -\beta }{2}}{\sin \left( \alpha -\beta \right)}$
$\begin{align}
  & x=\dfrac{2a\sin \dfrac{\alpha +\beta }{2}\cos \dfrac{\alpha -\beta }{2}}{2\sin \dfrac{\alpha -\beta }{2}\cos \dfrac{\alpha -\beta }{2}} \\
 & x=\dfrac{a\sin \dfrac{\alpha +\beta }{2}}{\sin \dfrac{\alpha -\beta }{2}} \\
\end{align}$
And,
$\begin{align}
  & y=\dfrac{2b\cos \dfrac{\alpha +\beta }{2}\cos \dfrac{\alpha -\beta }{2}}{\sin \left( \beta -\alpha \right)} \\
 & y=\dfrac{2b\cos \dfrac{\alpha +\beta }{2}\cos \dfrac{\alpha -\beta }{2}}{-2\sin \dfrac{\alpha -\beta }{2}\cos \dfrac{\alpha -\beta }{2}} \\
 & y=\dfrac{-b\cos \dfrac{\alpha +\beta }{2}}{\sin \dfrac{\alpha -\beta }{2}} \\
\end{align}$
Now, we have point $C\left( {{x}_{_{1}}},{{y}_{1}} \right)$as;
$\left( {{x}_{_{1}}},{{y}_{1}} \right)=\left( \dfrac{a\sin \dfrac{\alpha +\beta }{2}}{\sin \dfrac{\alpha -\beta }{2}},\dfrac{-b\cos \dfrac{\alpha +\beta }{2}}{\sin \dfrac{\alpha -\beta }{2}} \right).............\left( vii \right)$
As we have point $B(x,y)$ calculated in equation (iv) and will satisfy equation ${{x}^{2}}={{h}^{2}}$. Putting value of $x$ from equation (iv) to ${{x}^{2}}={{h}^{2}}$, we get;
$\dfrac{{{a}^{2}}{{\cos }^{2}}\dfrac{\alpha +\beta }{2}}{{{\cos }^{2}}\dfrac{\alpha -\beta }{2}}={{h}^{2}}...........\left( viii \right)$
Now, let us find out values of ${{\cos }^{2}}\dfrac{\alpha +\beta }{2}$,${{\cos }^{2}}\dfrac{\alpha -\beta }{2}$from equation (vii) in terms of ${{x}_{1}}\text{ and }{{y}_{1}}$, as;
${{x}_{_{1}}}=\dfrac{a\sin \dfrac{\alpha +\beta }{2}}{\sin \dfrac{\alpha -\beta }{2}},{{y}_{1}}=\dfrac{-b\cos \dfrac{\alpha +\beta }{2}}{\sin \dfrac{\alpha -\beta }{2}}$
Dividing ${{x}_{1}}\text{ and }{{y}_{1}}$, we get;
$\left( \dfrac{{{x}_{1}}}{{{y}_{1}}} \right)=\dfrac{-a}{b}\tan \left( \dfrac{\alpha +\beta }{2} \right)$
Squaring both sides, we get;
\[\left( \dfrac{{{x}_{1}}^{2}}{{{y}_{1}}^{2}} \right)=\dfrac{{{a}^{2}}}{{{b}^{2}}}{{\tan }^{2}}\left( \dfrac{\alpha +\beta }{2} \right)\]
We have relation $1+{{\tan }^{2}}\theta ={{\sec }^{2}}\theta $ or${{\tan }^{2}}\theta ={{\sec }^{2}}\theta -1,$using it in above equation, we get;
\[\begin{align}
  & \dfrac{{{b}^{2}}{{x}_{1}}^{2}}{{{a}^{2}}{{y}_{1}}^{2}}={{\sec }^{2}}\dfrac{\alpha +\beta }{2}-1 \\
 & {{\sec }^{2}}\dfrac{\alpha +\beta }{2}=1+\dfrac{{{b}^{2}}{{x}_{1}}^{2}}{{{a}^{2}}{{y}_{1}}^{2}} \\
 & {{\sec }^{2}}\dfrac{\alpha +\beta }{2}=\dfrac{{{a}^{2}}{{y}_{1}}^{2}+{{b}^{2}}{{x}_{1}}^{2}}{{{a}^{2}}{{y}_{1}}^{2}} \\
 & Or\text{ co}{{\text{s}}^{2}}\dfrac{\alpha +\beta }{2}=\dfrac{{{a}^{2}}{{y}_{1}}^{2}}{{{a}^{2}}{{y}_{1}}^{2}+{{b}^{2}}{{x}_{1}}^{2}}............\left( ix \right) \\
\end{align}\]
Now, let us calculate square of ${{x}_{1}}$;
${{x}_{_{1}}}^{2}=\dfrac{{{a}^{2}}{{\sin }^{2}}\dfrac{\alpha +\beta }{2}}{{{\sin }^{2}}\left( \dfrac{\alpha -\beta }{2} \right)}$
We have${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$; Hence;
$\begin{align}
  & {{x}_{_{1}}}^{2}=\dfrac{{{a}^{2}}\left( 1-{{\cos }^{2}}\dfrac{\alpha +\beta }{2} \right)}{{{\sin }^{2}}\left( \dfrac{\alpha -\beta }{2} \right)} \\
 & {{\sin }^{2}}\dfrac{\alpha -\beta }{2}=\dfrac{{{a}^{2}}}{{{x}_{_{1}}}^{2}}\left( 1-\dfrac{{{a}^{2}}{{y}_{1}}^{2}}{{{a}^{2}}{{y}_{1}}^{2}+{{b}^{2}}{{x}_{1}}^{2}} \right) \\
 & {{\sin }^{2}}\dfrac{\alpha -\beta }{2}=\dfrac{{{a}^{2}}}{{{x}_{_{1}}}^{2}}\left( \dfrac{{{b}^{2}}{{x}_{1}}^{2}}{{{a}^{2}}{{y}_{1}}^{2}+{{b}^{2}}{{x}_{1}}^{2}} \right) \\
 & {{\sin }^{2}}\dfrac{\alpha -\beta }{2}=\dfrac{{{a}^{2}}{{b}^{2}}}{{{a}^{2}}{{y}_{1}}^{2}+{{b}^{2}}{{x}_{1}}^{2}} \\
 & Or\text{ }1-{{\cos }^{2}}\dfrac{\alpha -\beta }{2}=\dfrac{{{a}^{2}}{{b}^{2}}}{{{a}^{2}}{{y}_{1}}^{2}+{{b}^{2}}{{x}_{1}}^{2}} \\
 & {{\cos }^{2}}\dfrac{\alpha -\beta }{2}=\dfrac{{{a}^{2}}{{y}_{1}}^{2}+{{b}^{2}}{{x}_{1}}^{2}-{{a}^{2}}{{b}^{2}}}{{{a}^{2}}{{y}_{1}}^{2}+{{b}^{2}}{{x}_{1}}^{2}}...........\left( x \right) \\
\end{align}$
Putting values of ${{\cos }^{2}}\dfrac{\alpha +\beta }{2},{{\cos }^{2}}\dfrac{\alpha -\beta }{2}$from equation (ix) and (x) in equation (viii), we get;
$\begin{align}
  & \dfrac{{{a}^{2}}\left( \dfrac{{{a}^{2}}{{y}_{1}}^{2}}{{{a}^{2}}{{y}_{1}}^{2}+{{b}^{2}}{{x}_{1}}^{2}} \right)}{\left( \dfrac{{{a}^{2}}{{y}_{1}}^{2}+{{b}^{2}}{{x}_{1}}^{2}-{{a}^{2}}{{b}^{2}}}{{{a}^{2}}{{y}_{1}}^{2}+{{b}^{2}}{{x}_{1}}^{2}} \right)}={{h}^{2}} \\
 & \dfrac{{{a}^{4}}{{y}_{1}}^{2}}{{{a}^{2}}{{y}_{1}}^{2}+{{b}^{2}}{{x}_{1}}^{2}-{{a}^{2}}{{b}^{2}}}={{h}^{2}} \\
 & \dfrac{{{a}^{4}}{{y}_{1}}^{2}}{{{h}^{2}}}={{a}^{2}}{{y}_{1}}^{2}+{{b}^{2}}{{x}_{1}}^{2}-{{a}^{2}}{{b}^{2}} \\
 & {{a}^{2}}{{b}^{2}}={{y}_{1}}^{2}\left( {{a}^{2}}-\dfrac{{{a}^{2}}}{{{h}^{2}}} \right)+{{b}^{2}}{{x}_{1}}^{2} \\
\end{align}$
Dividing both sides by ${{a}^{2}}{{b}^{2}}$, we get;
$1=\dfrac{{{y}_{1}}^{2}}{{{b}^{2}}}\left( 1-\dfrac{{{a}^{2}}}{{{h}^{2}}} \right)+\dfrac{{{x}_{1}}^{2}}{{{a}^{2}}}$
Replacing $\left( {{x}_{1}},{{y}_{1}} \right)$ by $\left( x,y \right)$ to get the required locus, we get;
$\dfrac{{{x}_{1}}^{2}}{{{a}^{2}}}+\dfrac{{{y}_{1}}^{2}}{{{b}^{2}}}\left( 1-\dfrac{{{a}^{2}}}{{{h}^{2}}} \right)=1$
Hence, proved.

Note: One can calculate intersection of tangent $PA$ and $AS$ to get point $A$ where point $P\text{ and }S$ has parametric coordinates as $\left( a\cos \alpha ,b\sin \alpha \right),\left( a\cos \left( \pi +\beta \right),b\sin \left( \pi +\beta \right) \right)$. Solution will be the same. Calculation is the important part of these kinds of questions.
One can use a substitution and elimination approach to find intersecting points of tangents but that will be a longer process than cross – multiplication which gives $\left( x,y \right)$ in one line.
By symmetry of ellipse points $\left( P,R \right)\text{ and }\left( Q,S \right)$ will be opposite to each other at a difference of $180{}^\circ $ eccentric angle which is the key point of this question..
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