Question

The line $\frac{x}{a}cos\theta -\frac{y}{b}sin\theta =1$ will touch ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at point P whose eccentric angle is

• A. $\theta$
• B. $(\pi -\theta )$
• C. $(\pi +\theta )$
• D. $2\pi -\theta$

Answer 1

The correct option is D $2\pi -\theta$

Given line equation $\frac{x}{a}cos\theta -\frac{y}{b}sin\theta =1andequationofellipseas\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Any point on the ellipse will be as $(acos\varphi ,bsin\varphi )$
Since line touches ellipse at the point P$(acos\varphi ,bsin\varphi )$
On substituting Point P on the line equation
$\Rightarrow \frac{acos\varphi }{a}cos\theta -\frac{bsin\varphi }{b}sin\theta =1$
$\Rightarrow cos\varphi cos\theta -sin\varphi sin\theta =1$
$\Rightarrow cos(\theta +\varphi )=cos2\pi$
$\Rightarrow (\theta +\varphi )=2\pi$
$\Rightarrow \varphi =2\pi -\theta$