Given:
- The equation of the straight line is passing through the point \( (a \cos^3 \theta, a \sin^3 \theta) \) and is perpendicular to the line \( x \sec \theta + y \csc \theta = a \).
To find the equation of the line that satisfies these conditions, we use the fact that the general form of the equation of a line perpendicular to the given line can be derived from the condition of perpendicularity.
The given line has the equation:
\[ x \sec \theta + y \csc \theta = a \]
The slope of this line is the negative reciprocal of the slope of the perpendicular line. To derive the equation of the perpendicular line, we use the point of intersection and the general formula for a line in the form \( Ax + By + C = 0 \).
After solving, the correct equation of the perpendicular line passing through the given point is:
\[ x \cos \theta - y \sin \theta = a \cos 2\theta \]
Therefore, the correct answer is (B): \( x \cos \theta - y \sin \theta = a \cos 2\theta \)
The given line is \(x \sec\theta + y \csc\theta = a\), which can be rewritten as \(\frac{x}{\cos\theta} + \frac{y}{\sin\theta} = a\). Multiplying by \(\sin\theta \cos\theta\), we get \(x \sin\theta + y \cos\theta = a \sin\theta \cos\theta\).
The slope of this line is \(m_1 = -\frac{\sin\theta}{\cos\theta} = -\tan\theta\).
The line we are looking for is perpendicular to this line. Therefore, its slope, \(m_2\), satisfies \(m_1 m_2 = -1\). Thus, \(m_2 = -\frac{1}{m_1} = \frac{\cos\theta}{\sin\theta} = \cot\theta\).
The line passes through the point \((a \cos^3\theta, a \sin^3\theta)\). Using the point-slope form of a line, \(y - y_1 = m(x - x_1)\), we have:
\(y - a \sin^3\theta = \cot\theta (x - a \cos^3\theta)\)
\(y - a \sin^3\theta = \frac{\cos\theta}{\sin\theta} (x - a \cos^3\theta)\)
\(y \sin\theta - a \sin^4\theta = x \cos\theta - a \cos^4\theta\)
\(x \cos\theta - y \sin\theta = a \cos^4\theta - a \sin^4\theta\)
\(x \cos\theta - y \sin\theta = a (\cos^4\theta - \sin^4\theta)\)
\(x \cos\theta - y \sin\theta = a (\cos^2\theta - \sin^2\theta)(\cos^2\theta + \sin^2\theta)\)
Since \(\cos^2\theta + \sin^2\theta = 1\), we have
\(x \cos\theta - y \sin\theta = a (\cos^2\theta - \sin^2\theta)\)
\(x \cos\theta - y \sin\theta = a \cos(2\theta)\)
Therefore, the equation of the line is \(x \cos\theta - y \sin\theta = a \cos(2\theta)\).
Let $ \mathbb{R} $ denote the set of all real numbers. Then the area of the region $$ \left\{ (x, y) \in \mathbb{R} \times \mathbb{R} : x > 0, y > \frac{1}{x},\ 5x - 4y - 1 > 0,\ 4x + 4y - 17 < 0 \right\} $$ is
Let $ S $ denote the locus of the point of intersection of the pair of lines $$ 4x - 3y = 12\alpha,\quad 4\alpha x + 3\alpha y = 12, $$ where $ \alpha $ varies over the set of non-zero real numbers. Let $ T $ be the tangent to $ S $ passing through the points $ (p, 0) $ and $ (0, q) $, $ q > 0 $, and parallel to the line $ 4x - \frac{3}{\sqrt{2}} y = 0 $.
Then the value of $ pq $ is