Find the locues of the middle point of the portion of the line `x"cos"alpha+y"sin"alpha=p`, where p is a costant, intercepted between the axes.

Answer

To find the locus of the midpoint of the portion of the line \( x \cos \alpha + y \sin \alpha = p \) that is intercepted between the axes, we will follow these steps:

### Step 1: Find the x-intercept and y-intercept of the line
1. **For the x-intercept**: Set \( y = 0 \) in the equation.
\[ x \cos \alpha + 0 \cdot \sin \alpha = p \implies x \cos \alpha = p \implies x = \frac{p}{\cos \alpha} \]
Thus, the x-intercept is \( \left( \frac{p}{\cos \alpha}, 0 \right) \).

2. **For the y-intercept**: Set \( x = 0 \) in the equation.
\[ 0 \cdot \cos \alpha + y \sin \alpha = p \implies y \sin \alpha = p \implies y = \frac{p}{\sin \alpha} \]
Thus, the y-intercept is \( \left( 0, \frac{p}{\sin \alpha} \right) \).

### Step 2: Find the midpoint of the line segment between the intercepts
The midpoint \( M \) of the line segment connecting the intercepts \( A \left( \frac{p}{\cos \alpha}, 0 \right) \) and \( B \left( 0, \frac{p}{\sin \alpha} \right) \) can be calculated as follows:
\[ M = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{\frac{p}{\cos \alpha} + 0}{2}, \frac{0 + \frac{p}{\sin \alpha}}{2} \right) \]
This simplifies to:
\[ M = \left( \frac{p}{2 \cos \alpha}, \frac{p}{2 \sin \alpha} \right) \]

### Step 3: Express the coordinates of the midpoint in terms of \( h \) and \( k \)
Let:
\[ h = \frac{p}{2 \cos \alpha}, \quad k = \frac{p}{2 \sin \alpha} \]
From these, we can express \( p \):
\[ p = 2h \cos \alpha \quad \text{and} \quad p = 2k \sin \alpha \]

### Step 4: Relate \( h \) and \( k \)
Equating the two expressions for \( p \):
\[ 2h \cos \alpha = 2k \sin \alpha \implies h \cos \alpha = k \sin \alpha \]
Rearranging gives:
\[ \frac{h}{k} = \frac{\sin \alpha}{\cos \alpha} = \tan \alpha \]

### Step 5: Use the identity \( \sin^2 \alpha + \cos^2 \alpha = 1 \)
From the relationship \( h \cos \alpha = k \sin \alpha \), we can square both sides:
\[ h^2 \cos^2 \alpha = k^2 \sin^2 \alpha \]
Substituting \( \cos^2 \alpha = 1 - \sin^2 \alpha \):
\[ h^2 (1 - \sin^2 \alpha) = k^2 \sin^2 \alpha \]
Rearranging gives:
\[ h^2 = k^2 \frac{\sin^2 \alpha}{1 - \sin^2 \alpha} \]

### Step 6: Substitute back to find the locus equation
Using the expressions for \( p \):
\[ h^2 + k^2 = \frac{p^2}{4} (\cos^2 \alpha + \sin^2 \alpha) = \frac{p^2}{4} \]
This leads to the equation:
\[ \frac{p^2}{4} = h^2 + k^2 \]
Thus, the locus of the midpoint is given by:
\[ \frac{p^2}{4h^2} + \frac{p^2}{4k^2} = 1 \]
This can be rearranged to:
\[ \frac{1}{h^2} + \frac{1}{k^2} = \frac{4}{p^2} \]

### Final Result
The locus of the midpoint of the line segment intercepted between the axes is:
\[ \frac{1}{h^2} + \frac{1}{k^2} = \frac{4}{p^2} \]