Question

Consider an ellipse $E: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Points Q and Q' are taken an opposite sides Q and Q'form a rectangle of maximum area. The eccentric angle of Q is

• A. $\tan^{-1} \sqrt{\frac{b}{a}}$
• B. $\tan^{-1} \frac{b}{a}$
• C. $2\tan^{-1} \sqrt{\frac{b}{a}}$
• D. $\tan^{-1} \sqrt{\frac{a}{b}}$

Answer 1

There seems to be a misunderstanding or a typo in the provided options or the question itself, as the standard interpretation of a rectangle of maximum area inscribed in an ellipse leads to an eccentric angle of $\frac{\pi}{4}$, which is not directly represented in the options for a general ellipse where $a \neq b$.

However, if the question is interpreted as finding the eccentric angle $\theta$ such that the area of the rectangle formed by the tangent at Q and the coordinate axes is MINIMIZED, then this occurs when $\theta = \frac{\pi}{4}$.

If the question is related to the area of the rectangle formed by the tangent at Q and the coordinate axes, and we are looking for the angle $\theta$ that MINIMIZES this area, then $\theta = \frac{\pi}{4}$.

A known result states that the area of the triangle formed by the tangent at Q and the coordinate axes is minimum when $\tan \theta = b/a$. The area of the rectangle formed by the tangent at Q and the coordinate axes is $A = \frac{ab}{|\sin\theta \cos\theta|}$. This area is minimized when $\theta = \pi/4$.

Given the options, and the commonality of certain related problems, it's possible the question is implicitly asking for an angle related to the tangent properties. A common problem is finding the angle $\theta$ such that the area of the triangle formed by the tangent at Q and the coordinate axes is minimum. This minimum occurs when $\tan \theta = b/a$. If this were the question, option (B) would be correct.

Another possibility is related to the angle subtended by the latus rectum at the center, or the angle related to the director circle.

Without further clarification or correction of the options, it is difficult to definitively select an answer. If forced to choose based on a common related problem (minimum area of triangle formed by tangent and axes), it would be related to $\tan \theta = b/a$. However, the question asks for maximum area of an inscribed rectangle.

Let's assume there's a specific context or a less common definition of "rectangle of maximum area" related to Q. If we consider the area of the rectangle formed by the tangent at Q and the coordinate axes, the area is $A = \frac{ab}{|\sin\theta \cos\theta|}$. To maximize this, $\sin\theta \cos\theta$ should be minimized, which occurs as $\theta \to 0$ or $\theta \to \pi/2$, leading to infinite area. To minimize this, $\sin\theta \cos\theta$ should be maximized, which occurs at $\theta = \pi/4$.

If the question implies the minimum area of the rectangle formed by the tangent at Q and the coordinate axes, then the eccentric angle is $\pi/4$. None of the options represent $\pi/4$ directly for a general ellipse.

However, a common problem asks for the angle $\theta$ such that the area of the triangle formed by the tangent at Q and the coordinate axes is minimum. This occurs when $\tan \theta = b/a$. If this were the intended question, option (B) would be correct.

Given the ambiguity and the mismatch of standard results with the options, it's impossible to provide a definitive correct answer. The standard problem of maximum area inscribed rectangle yields $\theta = \pi/4$. If the question is about minimum area of triangle formed by tangent and axes, it's $\tan \theta = b/a$.

Answer 2

The standard interpretation of finding a rectangle of maximum area inscribed in an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ leads to vertices at $(\pm \frac{a}{\sqrt{2}}, \pm \frac{b}{\sqrt{2}})$. The eccentric angle $\theta$ for these points is given by $a \cos \theta = \frac{a}{\sqrt{2}}$ and $b \sin \theta = \frac{b}{\sqrt{2}}$, which implies $\cos \theta = \frac{1}{\sqrt{2}}$ and $\sin \theta = \frac{1}{\sqrt{2}}$. Thus, $\theta = \frac{\pi}{4}$.

However, none of the options directly represent $\frac{\pi}{4}$ for a general ellipse where $a \neq b$. The options are: (A) $\tan^{-1} \sqrt{\frac{b}{a}}$ (B) $\tan^{-1} \frac{b}{a}$ (C) $2\tan^{-1} \sqrt{\frac{b}{a}}$ (D) $\tan^{-1} \sqrt{\frac{a}{b}}$

If the question intended to ask for the eccentric angle $\theta$ such that the area of the triangle formed by the tangent at Q and the coordinate axes is minimum, then this minimum occurs when $\tan \theta = \frac{b}{a}$. In this case, the answer would be $\theta = \tan^{-1} \frac{b}{a}$, which corresponds to option (B).

Given the discrepancy between the standard problem of maximum inscribed rectangle and the provided options, and considering a common related problem, option (B) might be the intended answer if the question was about minimizing the area of the triangle formed by the tangent and the axes. However, as stated, the question about maximum area of an inscribed rectangle does not yield any of the given options for a general ellipse.