Let a tangent be drawn to the ellipse

Question:

Let a tangent be drawn to the ellipse $\frac{x^{2}}{27}+y^{2}=1$ at $(3 \sqrt{3} \cos \theta, \sin \theta)$

where $\theta \in\left(0, \frac{\pi}{2}\right)$. Then the value of $\theta$ such that the sum of intercepts on axes made by this tangent is minimum is equal to:

 

  1. (1) $\frac{\pi}{8}$

     

  2. (2) $\frac{\pi}{4}$

  3. (3) $\frac{\pi}{6}$

  4. (4) $\frac{\pi}{3}$


Correct Option: 3,

Solution:

Equation of tangent be

$\frac{x \cos \theta}{3 \sqrt{3}}+\frac{y \cdot \sin \theta}{1}=1, \quad \theta \in\left(0, \frac{\pi}{2}\right)$

intercept on $\mathrm{x}$-axis

$\mathrm{OA}=3 \sqrt{3} \sec \theta$

intercept on $y$-axis

$\mathrm{OB}=\operatorname{cosec} \theta$

Now, sum of intercept

$=3 \sqrt{3} \sec \theta+\operatorname{cosec} \theta=f(\theta)$ let

$f^{\prime}(\theta)=3 \sqrt{3} \sec \theta \tan \theta-\operatorname{cosec} \theta \cot \theta$

$=3 \sqrt{3} \frac{\sin \theta}{\cos ^{2} \theta}-\frac{\cos \theta}{\sin ^{2} \theta}$

$=\frac{\cos \theta}{\sin ^{2} \theta} \cdot 3 \sqrt{3}\left[\tan ^{2} \theta-\frac{1}{3 \sqrt{3}}\right]=0 \Rightarrow \theta=\frac{\pi}{6}$

$\Rightarrow$ at $\theta=\frac{\pi}{6}, f(\theta)$ is minimum

 

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