Question

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

• A. \( 5a^2 \)
• B. \( 4a^2 \)
• C. \( 3a^2 \)
• D. \( 2a^2 \)
• E. \( a^2 \)

Hint:

This is a classic identity problem in coordinate geometry. The result is always the square of the constant term \( a \).

Answer 1

Answer: (E)

Concept: Distance \( d \) from origin to \( Ax + By - C = 0 \) is \( |C|/\sqrt{A^2 + B^2} \). We use the identities \( \sec\theta = 1/\cos\theta \), \( \csc\theta = 1/\sin\theta \), and \( \sin 2\theta = 2\sin\theta\cos\theta \).

Step 1: Calculate \( p \).
First line: \( \frac{x}{\cos\theta} + \frac{y}{\sin\theta} = a \Rightarrow x\sin\theta + y\cos\theta - a\sin\theta\cos\theta = 0 \). \[ p = \frac{|a\sin\theta\cos\theta|}{\sqrt{\sin^2\theta + \cos^2\theta}} = a\sin\theta\cos\theta = \frac{a}{2}\sin 2\theta \] \[ p^2 = \frac{a^2}{4}\sin^2 2\theta \quad \Rightarrow \quad 4p^2 = a^2\sin^2 2\theta \]

Step 2: Calculate \( q \).
Second line: \( x\cos\theta - y\sin\theta - a\cos 2\theta = 0 \). \[ q = \frac{|a\cos 2\theta|}{\sqrt{\cos^2\theta + \sin^2\theta}} = a\cos 2\theta \] \[ q^2 = a^2\cos^2 2\theta \]

Step 3: Sum \( 4p^2 + q^2 \).
\[ 4p^2 + q^2 = a^2\sin^2 2\theta + a^2\cos^2 2\theta \] \[ = a^2(\sin^2 2\theta + \cos^2 2\theta) = a^2(1) = a^2 \]