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. $\frac{\pi }{2}$
• C. ${\tan }^{-1}(1-e^2)$
• D. None of these

Answer 1

Answer: (B)

$\frac{\pi }{2}$

$P(\frac{a}{e},y^{\prime })=(\frac{a}{e},Y)$

$S(ae,0)$

$C(0,0)$

$PC:y=x\left(\frac{Y}{\frac{a}{e}}\right)$ meets ellipse at $A(x_1,y_1)$

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$\Rightarrow \frac{x^2}{a^2}+\frac{\frac{x^2{Y}^2}{a^2{e}^2}}{b^2}=1$

$\Rightarrow \frac{x^2}{a^2}+x^2{Y}^2\frac{(a^2-b^2)}{b^2}=1$

$\Rightarrow \frac{x^2}{a^2}+x^2{Y}^2\left(\frac{e^2}{1-e^2}\right)=1$

$x_1=\frac{1}{\sqrt{\frac{1}{a^2}+\frac{Y^2{e}^2}{1-e^2}}}$

Slope of tangent at $A=-\frac{b^2}{a^2}\frac{x_1}{y_1}=-\frac{b^2}{a^2}\times \frac{\frac{a}{e}}{Y}=(1-e^2)\times \frac{a}{eY}$

Slope of $PS=\frac{Y}{\frac{{ae}^2}{1-e^2}}=\frac{Ye}{a(1-e^2)}$

$\Rightarrow$ Product of slope of $PS$ and $T_A=-1$

$\alpha =\frac{\pi }{2}$