Locus of the mid points of the portion of the line `x cos theta+y sin theta =p` intercepted between the axes is ……………..

Answer

To find the locus of the midpoints of the portion of the line \( x \cos \theta + y \sin \theta = p \) intercepted between the axes, we can follow these steps:

### Step 1: Find the intercepts of the line with the axes
To find the intercepts of the line with the x-axis and y-axis, we set \( y = 0 \) and \( x = 0 \) respectively.

- **X-intercept**: Set \( y = 0 \):
\[ x \cos \theta = p \implies x = \frac{p}{\cos \theta} \]
So, the x-intercept is \( A\left(\frac{p}{\cos \theta}, 0\right) \).

- **Y-intercept**: Set \( x = 0 \):
\[ y \sin \theta = p \implies y = \frac{p}{\sin \theta} \]
So, the y-intercept is \( B\left(0, \frac{p}{\sin \theta}\right) \).

### Step 2: Find the midpoint of the segment AB
The midpoint \( P(h, k) \) of the segment \( AB \) can be calculated using the midpoint formula:
\[ h = \frac{x_1 + x_2}{2}, \quad k = \frac{y_1 + y_2}{2} \]
Substituting the coordinates of points \( A \) and \( B \):
\[ h = \frac{\frac{p}{\cos \theta} + 0}{2} = \frac{p}{2 \cos \theta} \]
\[ k = \frac{0 + \frac{p}{\sin \theta}}{2} = \frac{p}{2 \sin \theta} \]

### Step 3: Express \( h \) and \( k \) in terms of \( \theta \)
We have:
\[ h = \frac{p}{2 \cos \theta}, \quad k = \frac{p}{2 \sin \theta} \]

### Step 4: Find the relationship between \( h \) and \( k \)
To eliminate \( \theta \), we can express \( \frac{1}{h} \) and \( \frac{1}{k} \):
\[ \frac{1}{h} = \frac{2 \cos \theta}{p}, \quad \frac{1}{k} = \frac{2 \sin \theta}{p} \]

### Step 5: Square and add the equations
Now, squaring both sides and adding:
\[ \left(\frac{1}{h}\right)^2 + \left(\frac{1}{k}\right)^2 = \left(\frac{2 \cos \theta}{p}\right)^2 + \left(\frac{2 \sin \theta}{p}\right)^2 \]
\[ \frac{1}{h^2} + \frac{1}{k^2} = \frac{4 \cos^2 \theta + 4 \sin^2 \theta}{p^2} \]
Using the identity \( \cos^2 \theta + \sin^2 \theta = 1 \):
\[ \frac{1}{h^2} + \frac{1}{k^2} = \frac{4}{p^2} \]

### Step 6: Replace \( h \) and \( k \) with \( x \) and \( y \)
Since \( h = x \) and \( k = y \), we have:
\[ \frac{1}{x^2} + \frac{1}{y^2} = \frac{4}{p^2} \]

### Final Result
Rearranging gives the locus of the midpoints:
\[ \frac{1}{x^2} + \frac{1}{y^2} = \frac{4}{p^2} \]