Question

If \( p_1 \) and \( p_2 \) are respectively length of perpendiculars from the origin to the straight lines \( x\sec\theta + y\cosec\theta = a \) and \( x\cos\theta - y\sin\theta = a\cos 2\theta \), then \( 4p_1^2 + p_2^2 = \)

• A. \( 1 \)
• B. \( a^2 \)
• C. \( \frac{1}{a^2} \)
• D. \( a \)
• E. \( \frac{1}{a} \)

Hint:

Convert trigonometric expressions into standard identities like \( \sin^2\theta + \cos^2\theta = 1 \) to simplify quickly.

Answer 1

The Correct Option is B

Concept: Distance of a point \((x_0,y_0)\) from a line \(Ax+By+C=0\) is: \[ \text{Distance} = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \] For origin \((0,0)\), this simplifies to: \[ \frac{|C|}{\sqrt{A^2 + B^2}} \] We will apply this formula to both given lines.

Step 1: First line
Given: \[ x\sec\theta + y\cosec\theta = a \] Rewrite in standard form: \[ x\sec\theta + y\cosec\theta - a = 0 \] Thus: \[ A = \sec\theta,\quad B = \cosec\theta,\quad C = -a \]

Step 2: Distance \(p_1\)
\[ p_1 = \frac{|a|}{\sqrt{\sec^2\theta + \cosec^2\theta}} \] \[ p_1^2 = \frac{a^2}{\sec^2\theta + \cosec^2\theta} \]

Step 3: Simplify denominator
\[ \sec^2\theta = \frac{1}{\cos^2\theta}, \quad \cosec^2\theta = \frac{1}{\sin^2\theta} \] \[ \sec^2\theta + \cosec^2\theta = \frac{1}{\cos^2\theta} + \frac{1}{\sin^2\theta} = \frac{\sin^2\theta + \cos^2\theta}{\sin^2\theta \cos^2\theta} \] \[ = \frac{1}{\sin^2\theta \cos^2\theta} \]

Step 4: Therefore
\[ p_1^2 = a^2 \sin^2\theta \cos^2\theta \]

Step 5: Second line
\[ x\cos\theta - y\sin\theta = a\cos 2\theta \] Standard form: \[ x\cos\theta - y\sin\theta - a\cos 2\theta = 0 \] Thus: \[ A = \cos\theta,\quad B = -\sin\theta,\quad C = -a\cos 2\theta \]

Step 6: Distance \(p_2\)
\[ p_2 = \frac{|a\cos 2\theta|}{\sqrt{\cos^2\theta + \sin^2\theta}} \] \[ p_2 = |a\cos 2\theta| \] \[ p_2^2 = a^2 \cos^2 2\theta \]

Step 7: Compute expression
\[ 4p_1^2 + p_2^2 = 4(a^2 \sin^2\theta \cos^2\theta) + a^2 \cos^2 2\theta \] \[ = a^2 (4\sin^2\theta \cos^2\theta + \cos^2 2\theta) \]

Step 8: Use identity
\[ \sin 2\theta = 2\sin\theta \cos\theta \Rightarrow 4\sin^2\theta \cos^2\theta = \sin^2 2\theta \] Thus: \[ 4p_1^2 + p_2^2 = a^2(\sin^2 2\theta + \cos^2 2\theta) \]

Step 9: Apply identity
\[ \sin^2 2\theta + \cos^2 2\theta = 1 \]

Step 10: Final Answer
\[ 4p_1^2 + p_2^2 = a^2 \] \[ \boxed{a^2} \]