First slide

Ellipse

Question

Let P be any point on a directrix of an ellipse with 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 $α$ . Then $α$ is equal to

Difficult

A

$\tan^{- 1} e$
B

$\frac{π}{2}$
C

$\tan^{- 1} (1 - e^{2})$
D

None of these

Solution

$Equation of P C i s \frac{y}{x} = \frac{β}{\frac{a}{e}}$
$y = \frac{β e}{a} x point of intersection with PC is$ A with the ellipse
$A \equiv (\frac{a b}{\sqrt{b^{2} + β^{2} e^{2}}}, \frac{β e b}{\sqrt{b^{2} + β^{2} e^{2}}})$
$Step-III: Tangent at A \equiv \frac{a b x}{a^{2} {(b^{2} + β^{2} e^{2})}^{\frac{1}{2}}} + \frac{β e y}{b^{2} {(b^{2} + β^{2} e^{2})}^{\frac{1}{2}}} = 1$
$\Rightarrow \frac{b x}{a} + \frac{β e y}{b} = {(b^{2} + β^{2} e^{2})}^{\frac{1}{2}}$
$Equation of PS i s \frac{y}{x - a e} = \frac{β}{\frac{a}{e} - a e}$
$\tan α = \frac{m_{1} - m_{2}}{1 + m_{1} m_{2}}$
$Hence, m_{1} = \frac{- b^{2}}{a β e} m_{2} = \frac{β e}{a - a e^{2}}$
$Now, m_{1} \times m_{2} = \frac{- b^{2}}{a β e} \times \frac{β e}{a - a e^{2}}$
$= \frac{- b^{2}}{a^{2} (1 - e^{2})} = \frac{- b^{2}}{b^{2}} = - 1$
$Concept: So, m_{1} m_{2} + 1 = 0$
$\Rightarrow α = \frac{π}{2}$

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