Question

Let a tangent be drawn to the ellipse $\left(\frac{x^2}{27}\right)+y^2=1$ at $(3\sqrt{3}\cos \theta ,\sin \theta )$ where $\theta \in \left(0,\frac{\pi }{2}\right)$. Then the value of θ such that the sum of intercepts on axes made by a tangent is minimum is equal to:

• A. $\frac{\pi }{8}$
• B. $\frac{\pi }{6}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{4}$

Answer 1

The correct option is B

$\frac{\pi }{6}$

Explanation For The Correct Option :

Finding the value of $\theta$ such that the sum of intercepts on axes made by a tangent is minimum,

The given point is $(3\sqrt{3}\cos \theta ,\sin \theta )$

Equation of the tangent at this point:

$$\begin{gathered} \Rightarrow \frac{n\left(3\sqrt{3}\cos \theta \right)}{27}+\frac{y\left(\sin \theta \right)}{1}=1 \\ \Rightarrow \frac{n\cos \theta }{3}+y\sin \theta =1 \end{gathered}$$

Therefore, the sum of intersects on axes $=3\sqrt{3}sec\theta +csc\theta =f\left(\theta \right)$

$\Rightarrow f\text{'}\left(\theta \right)=\frac{3\sqrt{3}{\sin }^3\theta -{\cos }^3\theta }{{\sin }^2\theta .{\cos }^2\theta }...\left(i\right)$ [ Differentiating $f\left(\theta \right)$ w. r. to $\theta$]

For $f\left(\theta \right)$ to be minimum, $f\text{'}\left(\theta \right)=0$

$$\begin{gathered} \Rightarrow 3\sqrt{3}{\sin }^3\theta -{\cos }^3\theta =0[from(i\left)\right] \\ \Rightarrow 3\sqrt{3}{\sin }^3\theta ={\cos }^3\theta \\ \Rightarrow \tan \theta =\frac{1}{\sqrt{3}} \\ \Rightarrow \theta =\frac{\pi }{6}[taking{\tan }^{-1}bothsides] \end{gathered}$$

Hence, option B is the correct answer.