Key Concepts and Formulas

• Equation of a straight line in intercept form: A line intersecting the x-axis at $(a, 0)$ and the y-axis at $(0, b)$ has the equation $\frac{x}{a} + \frac{y}{b} = 1$.
• Midpoint Formula: The midpoint of a line segment connecting two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$.
• Trigonometric Identity: The fundamental identity $\cos^2 \alpha + \sin^2 \alpha = 1$ is crucial for eliminating the parameter $\alpha$.

Step-by-Step Solution

1. Understanding the Given Equation and Identifying the Goal

The equation of the line is given as $x \cos \alpha + y \sin \alpha = p$, where $p$ is a constant. We need to find the locus of the midpoint of the segment of this line intercepted between the x and y axes.

2. Finding the x-intercept

To find the x-intercept, we set $y = 0$ in the equation of the line: $x \cos \alpha + 0 \cdot \sin \alpha = p$ $x \cos \alpha = p$ $x = \frac{p}{\cos \alpha}$ So, the x-intercept is the point $A = \left(\frac{p}{\cos \alpha}, 0\right)$.

3. Finding the y-intercept

To find the y-intercept, we set $x = 0$ in the equation of the line: $0 \cdot \cos \alpha + y \sin \alpha = p$ $y \sin \alpha = p$ $y = \frac{p}{\sin \alpha}$ So, the y-intercept is the point $B = \left(0, \frac{p}{\sin \alpha}\right)$.

4. Finding the Midpoint of the Intercepted Segment

Let $M(h, k)$ be the midpoint of the line segment $AB$. Using the midpoint formula: $h = \frac{\frac{p}{\cos \alpha} + 0}{2} = \frac{p}{2 \cos \alpha}$ $k = \frac{0 + \frac{p}{\sin \alpha}}{2} = \frac{p}{2 \sin \alpha}$

5. Eliminating the Parameter $\alpha$

We want to find a relationship between $h$ and $k$ that doesn't involve $\alpha$. From the equations for $h$ and $k$, we have: $\cos \alpha = \frac{p}{2h}$ $\sin \alpha = \frac{p}{2k}$ Using the trigonometric identity $\cos^2 \alpha + \sin^2 \alpha = 1$, we substitute: $\left(\frac{p}{2h}\right)^2 + \left(\frac{p}{2k}\right)^2 = 1$ $\frac{p^2}{4h^2} + \frac{p^2}{4k^2} = 1$ Multiplying both sides by $\frac{4}{p^2}$, we get $\frac{1}{h^2} + \frac{1}{k^2} = \frac{4}{p^2}$

6. Stating the Locus

Replacing $(h, k)$ with $(x, y)$ to represent the general point on the locus, we get: $\frac{1}{x^2} + \frac{1}{y^2} = \frac{4}{p^2}$

Common Mistakes & Tips

• Be careful when isolating $\cos \alpha$ and $\sin \alpha$ from the midpoint equations.
• Remember to square both the numerator and the denominator when substituting into the trigonometric identity.
• Ensure that the final answer is in terms of $x$ and $y$, not $h$ and $k$.

Summary

We found the x and y intercepts of the given line, then found the midpoint of the segment between these intercepts. Finally, we eliminated the parameter $\alpha$ to find the equation of the locus of the midpoint, which is $\frac{1}{x^2} + \frac{1}{y^2} = \frac{4}{p^2}$.

The final answer is \boxed{D}, which corresponds to option (D).