The locus of the mid point of the portion intersection between the axes by the line `x cos alpha+ysin alpha=p` where p is constant is

Options

A. `x^(2)+y^(2)=4p^(2)`
B. `(1//x^(2))+(1//y^(2))=4//p^(2)`
C. `x^(2)+y^(2)=4//p^(2)`
D. `(1//x^(2))+(1//y^(2))=2//p^(2)`

Answer

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

### Step 1: Find the intersection points of the line with the axes
1. **Intersection with the x-axis**: Set \( y = 0 \) in the line equation:
\[ x \cos \alpha + 0 \cdot \sin \alpha = p \implies x = \frac{p}{\cos \alpha} \]
So, the intersection point with the x-axis is \( A\left(\frac{p}{\cos \alpha}, 0\right) \).

2. **Intersection with the y-axis**: Set \( x = 0 \) in the line equation:
\[ 0 \cdot \cos \alpha + y \sin \alpha = p \implies y = \frac{p}{\sin \alpha} \]
So, the intersection point with the y-axis is \( B\left(0, \frac{p}{\sin \alpha}\right) \).

### Step 2: Find the midpoint of the segment AB
The midpoint \( M(h, k) \) of the segment joining points \( A \) and \( B \) is given by:
\[ h = \frac{x_1 + x_2}{2} = \frac{\frac{p}{\cos \alpha} + 0}{2} = \frac{p}{2 \cos \alpha} \]
\[ k = \frac{y_1 + y_2}{2} = \frac{0 + \frac{p}{\sin \alpha}}{2} = \frac{p}{2 \sin \alpha} \]

### Step 3: Express \( \frac{1}{h} \) and \( \frac{1}{k} \)
From the midpoint coordinates, we have:
\[ \frac{1}{h} = \frac{2 \cos \alpha}{p} \quad \text{and} \quad \frac{1}{k} = \frac{2 \sin \alpha}{p} \]

### Step 4: Square and add the equations
Now, we square both equations and add them:
\[ \left(\frac{1}{h}\right)^2 + \left(\frac{1}{k}\right)^2 = \left(\frac{2 \cos \alpha}{p}\right)^2 + \left(\frac{2 \sin \alpha}{p}\right)^2 \]
\[ \frac{1}{h^2} + \frac{1}{k^2} = \frac{4 \cos^2 \alpha}{p^2} + \frac{4 \sin^2 \alpha}{p^2} \]
\[ = \frac{4(\cos^2 \alpha + \sin^2 \alpha)}{p^2} = \frac{4}{p^2} \]

### Step 5: Substitute \( h \) and \( k \) back
Substituting \( h = x \) and \( k = y \):
\[ \frac{1}{x^2} + \frac{1}{y^2} = \frac{4}{p^2} \]

### Step 6: Rearranging the equation
Multiplying through by \( x^2y^2p^2 \):
\[ y^2p^2 + x^2p^2 = 4x^2y^2 \]
This is the equation of the locus.

### Final Result
The locus of the midpoint of the portion of the line that intersects the axes is given by:
\[ \frac{1}{x^2} + \frac{1}{y^2} = \frac{4}{p^2} \]