Question

Let a tangent be drawn to the ellipse x²/27 + y² = 1 at (3$\sqrt{3}$ cosθ, sinθ) where θ ∈ (0, π/2). Then the value of θ such that the sum of intercepts on axes made by this tangent is minimum is equal to :

• A. π/3
• B. π/6
• C. π/8
• D. π/4

Hint:

To minimize $a \sec \theta + b \csc \theta$, set $\tan \theta = (b/a)^{1/3}$.

Answer 1

The Correct Option is B

Step 1: Equation of tangent at $(a \cos \theta, b \sin \theta)$ is $\frac{x \cos \theta}{a} + \frac{y \sin \theta}{b} = 1$. Here $a=3\sqrt{3}, b=1$. Tangent: $\frac{x \cos \theta}{3\sqrt{3}} + y \sin \theta = 1$.
Step 2: Intercepts are $x_{int} = \frac{3\sqrt{3}}{\cos \theta}$ and $y_{int} = \frac{1}{\sin \theta}$. Sum $S = 3\sqrt{3} \sec \theta + \csc \theta$.
Step 3: For minimum $S$, $\frac{dS}{d\theta} = 3\sqrt{3} \sec \theta \tan \theta - \csc \theta \cot \theta = 0$. $\frac{3\sqrt{3} \sin \theta}{\cos^2 \theta} = \frac{\cos \theta}{\sin^2 \theta} \implies \tan^3 \theta = \frac{1}{3\sqrt{3}} = (\frac{1}{\sqrt{3}})^3$. $\tan \theta = \frac{1}{\sqrt{3}} \implies \theta = \pi/6$.