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
$\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}\left( 1-\dfrac{{{a}^{2}}}{{{h}^{2}}} \right)=1$.
Last updated date: 21st Mar 2023
•
Total views: 306.6k
•
Views today: 2.85k
Answer
306.6k+ views
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..
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..
Recently Updated Pages
Calculate the entropy change involved in the conversion class 11 chemistry JEE_Main

The law formulated by Dr Nernst is A First law of thermodynamics class 11 chemistry JEE_Main

For the reaction at rm0rm0rmC and normal pressure A class 11 chemistry JEE_Main

An engine operating between rm15rm0rm0rmCand rm2rm5rm0rmC class 11 chemistry JEE_Main

For the reaction rm2Clg to rmCrmlrm2rmg the signs of class 11 chemistry JEE_Main

The enthalpy change for the transition of liquid water class 11 chemistry JEE_Main

Trending doubts
Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

List out three methods of soil conservation

Fill in the blanks A 1 lakh ten thousand B 1 million class 9 maths CBSE

Write a letter to the Principal of your school to plead class 10 english CBSE
