1. Write the given line in intercept form

The line is $x\cos\alpha + y\sin\alpha = p.$

Its intercepts on the axes are found by setting one variable to zero.

• On the $x$-axis: put $y=0$ $x\cos\alpha = p \implies x = p\sec\alpha.$ So the point is $A(p\sec\alpha,0)$.

• On the $y$-axis: put $x=0$ $y\sin\alpha = p \implies y = p\csc\alpha.$ So the point is $B(0,p\csc\alpha)$.

2. Find the midpoint of the intercepted portion

Let the midpoint be $M(x,y)$.

Using midpoint formula for $A(p\sec\alpha,0)$ and $B(0,p\csc\alpha)$, $x = \frac{p\sec\alpha}{2}, \qquad y = \frac{p\csc\alpha}{2}.$

So, $\sec\alpha = \frac{2x}{p}, \qquad \csc\alpha = \frac{2y}{p}.$

3. Eliminate the parameter $\alpha$

Convert to cosine and sine: $\cos\alpha = \frac{p}{2x}, \qquad \sin\alpha = \frac{p}{2y}.$

Now use the identity $\sin^2\alpha + \cos^2\alpha = 1.$

Hence, $\left(\frac{p}{2x}\right)^2 + \left(\frac{p}{2y}\right)^2 = 1.$

Multiply through: $\frac{p^2}{4x^2} + \frac{p^2}{4y^2} = 1.$

Therefore, $\frac{1}{x^2} + \frac{1}{y^2} = \frac{4}{p^2}.$

4. Match with the options

This corresponds to $\frac{1}{x^2} + \frac{1}{y^2} = \frac{4}{p^2},$ which is Option D.

5. Comparison with stored answer

Stored correct answer: D

Derived answer: D

They match.