If p and q are respectively the perpendiculars from the origin upon the striaght lines, whose equations are ` x sec theta + y cosec theta =a and x cos theta -y sin theta = acos 2 theta , then 4p^(2) + q^(2)` is equal to

Options

A. `p^(2) +p'^(2)=a^(2)`
B. `p^(2)+4p'^(2)=a^(2)`
C. `4p^(2) +p'^(2)=a^(2)`
D. `4p^(2) +p'^(2)=4a^(2)`

Explanation

To solve the problem, we need to find the value of \(4p^2 + q^2\), where \(p\) and \(q\) are the perpendicular distances from the origin to the given lines.

### Step 1: Find the perpendicular distance \(p\) from the origin to the first line
The equation of the first line is:
\[ x \sec \theta + y \csc \theta = a \]
The formula for the perpendicular distance \(d\) from a point \((x_0, y_0)\) to the line \(Ax + By + C = 0\) is:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
For the first line, we can rewrite it as:
\[ x \sec \theta + y \csc \theta - a = 0 \]
Here, \(A = \sec \theta\), \(B = \csc \theta\), and \(C = -a\). Substituting \(x_0 = 0\) and \(y_0 = 0\) into the formula gives:
\[ p = \frac{|0 + 0 - a|}{\sqrt{(\sec \theta)^2 + (\csc \theta)^2}} = \frac{a}{\sqrt{\sec^2 \theta + \csc^2 \theta}} \]

### Step 2: Simplify the denominator
We know:
\[ \sec^2 \theta = \frac{1}{\cos^2 \theta}, \quad \csc^2 \theta = \frac{1}{\sin^2 \theta} \]
Thus,
\[ \sec^2 \theta + \csc^2 \theta = \frac{1}{\cos^2 \theta} + \frac{1}{\sin^2 \theta} = \frac{\sin^2 \theta + \cos^2 \theta}{\sin^2 \theta \cos^2 \theta} = \frac{1}{\sin^2 \theta \cos^2 \theta} \]
So,
\[ p = \frac{a}{\sqrt{\frac{1}{\sin^2 \theta \cos^2 \theta}}} = a \sin \theta \cos \theta \]

### Step 3: Express \(p\) in terms of \(a\) and \(\theta\)
Using the double angle identity:
\[ \sin 2\theta = 2 \sin \theta \cos \theta \]
We can write:
\[ p = \frac{a}{2} \sin 2\theta \]

### Step 4: Calculate \(4p^2\)
Now, we find \(4p^2\):
\[ 4p^2 = 4\left(\frac{a}{2} \sin 2\theta\right)^2 = a^2 \sin^2 2\theta \]

### Step 5: Find the perpendicular distance \(q\) from the origin to the second line
The equation of the second line is:
\[ x \cos \theta - y \sin \theta = a \cos 2\theta \]
Rewriting it gives:
\[ x \cos \theta - y \sin \theta - a \cos 2\theta = 0 \]
Using the same formula for \(q\):
\[ q = \frac{|0 - 0 - a \cos 2\theta|}{\sqrt{(\cos \theta)^2 + (-\sin \theta)^2}} = \frac{a \cos 2\theta}{\sqrt{\cos^2 \theta + \sin^2 \theta}} = a \cos 2\theta \]

### Step 6: Calculate \(q^2\)
Now we find \(q^2\):
\[ q^2 = (a \cos 2\theta)^2 = a^2 \cos^2 2\theta \]

### Step 7: Combine \(4p^2\) and \(q^2\)
Now we can find \(4p^2 + q^2\):
\[ 4p^2 + q^2 = a^2 \sin^2 2\theta + a^2 \cos^2 2\theta \]
Factoring out \(a^2\):
\[ 4p^2 + q^2 = a^2 (\sin^2 2\theta + \cos^2 2\theta) \]
Using the Pythagorean identity:
\[ \sin^2 2\theta + \cos^2 2\theta = 1 \]
Thus,
\[ 4p^2 + q^2 = a^2 \cdot 1 = a^2 \]

### Final Answer
Therefore, the value of \(4p^2 + q^2\) is:
\[ \boxed{a^2} \]