Distance of the lines `xcosectheta+ysectheta=kcot2theta` and `xsintheta+ycostheta=ksin2theta` from origin is p and q respectively then

Options

A. `4q^2+p^2=k^2`
B. `p^2+q^2=4k^2`
C. `4p^2+q^2=k^2`
D. `4p^2+q^2=4k^2`

Answer

To solve the problem, we need to find the distances \( p \) and \( q \) from the origin to the given lines and then derive the relationship between these distances and \( k \).

### Step 1: Write the equations of the lines
The two lines given are:
1. \( x \csc \theta + y \sec \theta = k \cot 2\theta \)
2. \( x \sin \theta + y \cos \theta = k \sin 2\theta \)

### Step 2: Convert the first line into standard form
The first line can be rewritten using the definitions of cosecant and secant:
\[ \frac{x}{\sin \theta} + \frac{y}{\cos \theta} = k \cdot \frac{\cos 2\theta}{\sin 2\theta} \]
This can be rearranged to:
\[ x \cos \theta + y \sin \theta = \frac{k}{2} \cos 2\theta \]

### Step 3: Find the distance \( p \) from the origin to the first line
The distance \( p \) from the origin (0,0) to the line \( Ax + By + C = 0 \) is given by:
\[ p = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} \]
For our line:
- \( A = \cos \theta \)
- \( B = \sin \theta \)
- \( C = -\frac{k}{2} \cos 2\theta \)

Thus,
\[ p = \frac{|0 + 0 - \frac{k}{2} \cos 2\theta|}{\sqrt{\cos^2 \theta + \sin^2 \theta}} = \frac{\frac{k}{2} |\cos 2\theta|}{1} = \frac{k}{2} |\cos 2\theta| \]

### Step 4: Convert the second line into standard form
The second line can be rearranged as:
\[ x \sin \theta + y \cos \theta = k \sin 2\theta \]

### Step 5: Find the distance \( q \) from the origin to the second line
Using the same formula for distance:
- \( A = \sin \theta \)
- \( B = \cos \theta \)
- \( C = -k \sin 2\theta \)

Thus,
\[ q = \frac{|0 + 0 - k \sin 2\theta|}{\sqrt{\sin^2 \theta + \cos^2 \theta}} = \frac{k |\sin 2\theta|}{1} = k |\sin 2\theta| \]

### Step 6: Relate \( \cos 2\theta \) and \( \sin 2\theta \) to \( p \) and \( q \)
From the distances calculated:
\[ \cos 2\theta = \frac{2p}{k} \quad \text{and} \quad \sin 2\theta = \frac{q}{k} \]

### Step 7: Use the Pythagorean identity
Using the identity \( \cos^2 2\theta + \sin^2 2\theta = 1 \):
\[ \left(\frac{2p}{k}\right)^2 + \left(\frac{q}{k}\right)^2 = 1 \]
This simplifies to:
\[ \frac{4p^2}{k^2} + \frac{q^2}{k^2} = 1 \]
Multiplying through by \( k^2 \) gives:
\[ 4p^2 + q^2 = k^2 \]

### Final Answer
Thus, the relationship is:
\[ \boxed{4p^2 + q^2 = k^2} \]