Question

If \(\beta\) is one of the angles between the normals to the ellipse, \(x^2 + 3y^2 = 9\) at the points \((3\cos\theta, \sqrt3\sin\theta)\) and \((-3\sin\theta, \sqrt3\cos\theta)\); \(\theta \in \left(0, \frac{\pi}{2}\right)\); then \(\frac{2\cot\beta}{\sin2\theta}\) is equal to :

• A. \(\frac{2}{\sqrt3}\)
• B. \(\frac{1}{\sqrt3}\)
• C. \(\sqrt2\)
• D. \(\frac{\sqrt3}{4}\)

Hint:

Normal direction at a point of x^2 + 3y^2 = 9 is the gradient (x, 3y); find the angle between the two normal vectors.

Answer 1

The Correct Option is A

The ellipse is \(\frac{x^2}{9}+\frac{y^2}{3}=1\). The normal at \((3\cos\theta,\sqrt3\sin\theta)\) is along the gradient \((x, 3y) = (3\cos\theta, 3\sqrt3\sin\theta)\), i.e. direction \(\vec n_1 = (\cos\theta, \sqrt3\sin\theta)\).
At \((-3\sin\theta, \sqrt3\cos\theta)\): \(\vec n_2 = (-\sin\theta, \sqrt3\cos\theta)\).
\(\vec n_1\cdot\vec n_2 = -\sin\theta\cos\theta + 3\sin\theta\cos\theta = 2\sin\theta\cos\theta = \sin2\theta\).
\(|\vec n_1|^2|\vec n_2|^2 = (1+2\sin^2\theta)(1+2\cos^2\theta) = 3 + \sin^2 2\theta\).
\(\cos\beta = \frac{\sin2\theta}{\sqrt{3+\sin^22\theta}}\), so \(\sin\beta = \frac{\sqrt3}{\sqrt{3+\sin^22\theta}}\) and \(\cot\beta = \frac{\sin2\theta}{\sqrt3}\).
Hence \(\frac{2\cot\beta}{\sin2\theta} = \frac{2}{\sqrt3}\).
Answer: (1) \(\frac{2}{\sqrt3}\)