Question

The parametric angle $\theta$, where $-\pi < \theta \le \pi$, of the point on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at which the tangent drawn cuts the intercept of minimum length on the co-ordinates axis, is/are :

• A. $\tan^{-1}\sqrt{\frac{b}{a}}$
• B. $-\tan^{-1}\sqrt{\frac{b}{a}}$
• C. $\pi-\tan^{-1}\sqrt{\frac{b}{a}}$
• D. $\pi+\tan^{-1}\sqrt{\frac{b}{a}}$

Answer 1

Answer: (A), (B), (C)

A, B, C

Answer 2

The parametric coordinates of a point on the ellipse are $(a \cos\theta, b \sin\theta)$. The equation of the tangent is $\frac{x}{a}\cos\theta + \frac{y}{b}\sin\theta = 1$. The intercepts on the axes are $X = \frac{a}{\cos\theta}$ and $Y = \frac{b}{\sin\theta}$. The square of the length of the intercept is $S^2 = X^2 + Y^2 = \frac{a^2}{\cos^2\theta} + \frac{b^2}{\sin^2\theta}$. Minimizing $S^2$ with respect to $\theta$ yields $\tan^2\theta = \frac{b}{a}$. Let $\alpha = \tan^{-1}\sqrt{\frac{b}{a}}$. Since $\sqrt{\frac{b}{a}} > 0$, $\alpha \in (0, \frac{\pi}{2})$. The condition $\tan^2\theta = \frac{b}{a}$ implies $\tan\theta = \pm \sqrt{\frac{b}{a}}$. The solutions for $\theta$ in the range $(-\pi, \pi]$ are:

1. $\tan\theta = \sqrt{\frac{b}{a}} = \tan\alpha \implies \theta = \alpha$ (Option A) or $\theta = \alpha - \pi$ (not an option).
2. $\tan\theta = -\sqrt{\frac{b}{a}} = -\tan\alpha = \tan(-\alpha) \implies \theta = -\alpha$ (Option B) or $\theta = \pi - \alpha$ (Option C).

The four angles in $(-\pi, \pi]$ are $\alpha$, $-\alpha$, $\pi-\alpha$, and $\alpha-\pi$. Options A, B, and C correspond to $\alpha$, $-\alpha$, and $\pi-\alpha$. Option D, $\pi+\alpha$, lies outside the range $(-\pi, \pi]$.