Question

Consider the curve $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. The portion of the tangent at any point of the curve intercepted between the point of contact and the directrix subtends at the corresponding focus an angle of

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{3}$
• C. $\frac{\pi }{2}$
• D. $\frac{\pi }{6}$

Answer 1

The correct option is C $\frac{\pi }{2}$


Equation of tangent at any point $P(a\cos \theta ,b\sin \theta )$ on the ellipse is $\frac{x\cos \theta }{a}+\frac{y\sin \theta }{b}=1$

When $x=\frac{a}{e},y=(1-\frac{\cos \theta }{e})\frac{b}{\sin \theta }$

$\therefore Q\equiv (\frac{a}{e},(1-\frac{\cos \theta }{e})\frac{b}{\sin \theta })$

Now, Slope of $SQ\times$ Slope of $PS$
$=\frac{(1-\frac{\cos \theta }{e})\frac{b}{\sin \theta }}{\frac{a}{e}-ae}\times \frac{b\sin \theta }{a\cos \theta -ae}$

$=\frac{(e-\cos \theta )b}{a(1-e^2)\sin \theta }\times \frac{b\sin \theta }{a(\cos \theta -e)}$

$=\frac{-b^2}{a^2(1-e^2)}$
$=\frac{-b^2\cdot a^2}{a^2\cdot b^2}$
$=-1$

So, the angle between the line $PS$ and $SQ$ is $\frac{\pi }{2}$.