Question

Let $P$ be any point on a directrix of an ellipse of eccentricity $e$ . $S$ be the corresponding focus and $C$ the centre of the ellipse. The line $PC$ meets the ellipse at $A$ . The angle between $PS$ and tangent at $A$ is $\alpha$ , then $\alpha$ is equal to
a. ${\tan }^{-1}e$
b. ${\displaystyle \frac{\pi }{2}}$
c. ${\tan }^{-1}\left(1-e^2\right)$
d.None of these

Answer 1

Hint: The point $P$ is equal to $\left({\displaystyle \frac{a}{e}},Y\right)$ , since the point $y$ meets in ellipse so $y$ is equal to the point $x\left({\displaystyle \frac{{\displaystyle \frac{Y}{a}}}{e}}\right)$ . Then substitute $y$ in the equation of ellipse to find the tangent at $A$ . Then we will determine slope in $PS$ . Product of the lope $PS$ and $A$ is equal to $-1$ which will help to determine the value of $\alpha$ .

Complete step-by-step answer:
The following is the schematic diagram of the ellipse in which $S$ is the corresponding focus and $C$ is the centre of the ellipse.

From the above diagram we observe that the point $A$ is $\left(a\cos \theta ,b\sin \theta \right)$ which is $\left(x_1,y_1\right)$ . The point $S$ is in the $S\left(ae,0\right)$ and the point $C$ is $\left(0,0\right)$ .
Equation of ellipse is ${\displaystyle \frac{x^2}{a^2}}+{\displaystyle \frac{y^2}{b^2}}=1$ .
Now, let the point $P$ is in the outer part of ellipse,
 $P\left({\displaystyle \frac{a}{e}},Y^{\prime }\right)=\left({\displaystyle \frac{a}{e}},Y\right)$
Since we know that the point $y$ meets at ellipse at $A$ that is at $\left(x_1,y_1\right)$ we get,
 $y=x\left({\displaystyle \frac{{\displaystyle \frac{Y}{a}}}{e}}\right)$
Now, we know that the equation of ellipse is,
 ${\displaystyle \frac{x^2}{a^2}}+{\displaystyle \frac{y^2}{b^2}}=1$
Since $y$ lies in the ellipse so the equation changes to,
 $\begin{array}{c}{\displaystyle \frac{x^2}{a^2}}+{\displaystyle \frac{y^2}{b^2}}=1\\ {\displaystyle \frac{x^2}{a^2}}+{\displaystyle \frac{{\displaystyle \frac{x^2{Y}^2}{a^2{e}^2}}}{b^2}}=1\end{array}$
On further solving the above expression, we get the value as,
 ${\displaystyle \frac{x^2}{a^2}}+x^2{Y}^2{\displaystyle \frac{\left(a^2-b^2\right)}{b^2}}=1$
Since, the eccentricity $e$ is equal to $\sqrt{a^2-b^2}$ . So, let us substitute the value we obtain,
 $\begin{array}{l}{\displaystyle \frac{x^2}{a^2}}+x^2{Y}^2{\displaystyle \frac{e^2}{1-e^2}}=1\\ x^2\left({\displaystyle \frac{1}{a^2}}+{\displaystyle \frac{Y^2{e}^2}{1-e^2}}\right)=1\end{array}$
The take term $\left({\displaystyle \frac{1}{a^2}}+{\displaystyle \frac{Y^2{e}^2}{1-e^2}}\right)$ to the right side and then take the square root both sides then we get,
 $\begin{array}{l}{x}^2={\displaystyle \frac{1}{\left({\displaystyle \frac{1}{a^2}}+{\displaystyle \frac{Y^2{e}^2}{1-e^2}}\right)}}\\ x={\displaystyle \frac{1}{\sqrt{\left({\displaystyle \frac{1}{a^2}}+{\displaystyle \frac{Y^2{e}^2}{1-e^2}}\right)}}}\end{array}$
This implies that $x={\displaystyle \frac{1}{\sqrt{\left({\displaystyle \frac{1}{a^2}}+{\displaystyle \frac{Y^2{e}^2}{1-e^2}}\right)}}}$ .
Now, we have to find the slope of the tangent at the point $A$ is equal to $-{\displaystyle \frac{b^2}{a^2}}{\displaystyle \frac{x_1}{y_1}}$ .
Since, we know that $y=x\left({\displaystyle \frac{{\displaystyle \frac{Y}{a}}}{e}}\right)$ , let us substitute in the above equation, so we get,
 $\begin{array}{c}{T}_{\mathrm{A}}=-{\displaystyle \frac{b^2}{a^2}}\times {\displaystyle \frac{{\displaystyle \frac{a}{e}}}{Y}}\\ =-\left(1-e^2\right)\times {\displaystyle \frac{a}{eY}}\end{array}$
Also, slope of $PS$ is equal to,
  ${\displaystyle \frac{Y}{{\displaystyle \frac{a{e}^2}{1-e^2}}}}={\displaystyle \frac{Ye}{a\left(1-e^2\right)}}$
Now, we will calculate the product of slope of $PS$ and $T_A$ which is given as,
  $\begin{array}{l}=\left[-\left(1-e^2\right)\times {\displaystyle \frac{a}{eY}}\right]\times \left[{\displaystyle \frac{Ye}{a\left(1-e^2\right)}}\right]\\ =-1\end{array}$

Then, we can say that $\alpha ={\displaystyle \frac{\pi }{2}}$ because PS is perpendicular to the tangent.
Hence, the correct option is ${\displaystyle \frac{\pi }{2}}$ .
So, the correct answer is “Option b”.

Note: Do not forget to take the $y$ at the $x\left({\displaystyle \frac{{\displaystyle \frac{Y}{a}}}{e}}\right)$ and this can. also be done by different methods. Also, take $A$ as $\left(a\cos \theta ,b\sin \theta \right)$ and equation of $AC$ is $y={\displaystyle \frac{b}{a}}x\tan \theta$ where, $\tan \theta$ is the slope.