To find the locus of midpoints, first find the general coordinates of the midpoint in terms of a parameter (like \( \theta \) in this case). Then, eliminate the parameter to get the equation of the locus.
Let's analyze the problem step by step and find out the value of \(\alpha\).
The line equation given is \(x \cos \theta + y \sin \theta = 7\), where \(\theta \in (0, \frac{\pi}{2})\). This represents a line in the first quadrant making angles with x-axis and y-axis depending on the value of \(\theta\).
The line segment discussed is between the x-axis and y-axis. For such a line segment, the x-intercept is obtained by setting \(y=0\):
\(x \cos \theta = 7 \Rightarrow x = \frac{7}{\cos \theta}\)
Similarly, the y-intercept is obtained by setting \(x=0\):
\(y \sin \theta = 7 \Rightarrow y = \frac{7}{\sin \theta}\)
The mid-point of the line segment joining these intercepts on the coordinate axes is found by averaging their coordinates. Thus, the mid-point \((x_m, y_m)\) is:
\(x_m = \frac{1}{2}\left(0 + \frac{7}{\cos \theta}\right) = \frac{7}{2\cos \theta}\)
\(y_m = \frac{1}{2}\left(\frac{7}{\sin \theta} + 0\right) = \frac{7}{2\sin \theta}\)
According to the problem, this point \((x_m, y_m)\) is given as \((\alpha, \frac{7\sqrt{3}}{3})\).
Equating the y-coordinate:
\(\frac{7}{2\sin \theta} = \frac{7\sqrt{3}}{3}\)
Solving for \(\sin \theta\):
\(\frac{1}{2\sin \theta} = \frac{\sqrt{3}}{3} \Rightarrow 2\sin \theta = \frac{3}{\sqrt{3}} = \sqrt{3} \Rightarrow \sin \theta = \frac{\sqrt{3}}{2}\)
The angle \(\theta\) which satisfies this is \(\theta = \frac{\pi}{3}\) as \(\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}\).
Now, evaluating \(x_m\) when \(\cos \theta\) at this angle:
\(\cos \theta = \frac{1}{2}\) when \(\theta = \frac{\pi}{3}\).
Thus, \(x_m = \frac{7}{2 \times \frac{1}{2}} = 7\).
The value of \(\alpha\) from \((\alpha, \frac{7\sqrt{3}}{3})\) becomes 7, which matches one of the given options.
Thus, the correct answer is:
Option: 7
The area enclosed by the closed curve $C$ given by the differential equation $\frac{d y}{d x}+\frac{x+a}{y-2}=0, y(1)=0$ is $4 \pi$.
Let $P$ and $Q$ be the points of intersection of the curve $C$ and the $y$-axis If normals at $P$ and $Q$ on the curve $C$ intersect $x$-axis at points $R$ and $S$ respectively, then the length of the line segment $R S$ is