Question

If \(p\) and \(q\) are the lengths of the perpendiculars from the origin on the lines, \(x \csc \alpha - y \sec \alpha = k \cot 2\alpha\) and \(x \sin \alpha + y \cos \alpha = k \sin 2\alpha\) respectively, then \(k^2\) is equal to :

• A. \(p^2 + 2q^2\)
• B. \(p^2 + 4q^2\)
• C. \(2p^2 + q^2\)
• D. \(4p^2 + q^2\)

Hint:

Simplify trigonometric coefficients into sine and cosine forms before calculating distances. It usually leads to the identity \(\sin^2 \theta + \cos^2 \theta = 1\).

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The length of the perpendicular from the origin \((0, 0)\) to the line \(Ax + By + C = 0\) is given by \(d = \frac{|C|}{\sqrt{A^2 + B^2}}\).
Step 2: Detailed Explanation:
Line 1: \(x \csc \alpha - y \sec \alpha = k \cot 2\alpha \implies \frac{x}{\sin \alpha} - \frac{y}{\cos \alpha} = \frac{k \cos 2\alpha}{\sin 2\alpha}\).
Multiply by \(\sin \alpha \cos \alpha\):
\[ x \cos \alpha - y \sin \alpha = \frac{k \cos 2\alpha \cdot \sin \alpha \cos \alpha}{\sin 2\alpha} = \frac{k \cos 2\alpha \cdot \sin \alpha \cos \alpha}{2 \sin \alpha \cos \alpha} = \frac{k}{2} \cos 2\alpha \]
So, \(p = \frac{|-(k/2) \cos 2\alpha|}{\sqrt{\cos^2 \alpha + \sin^2 \alpha}} = \frac{k}{2} |\cos 2\alpha| \implies 2p = k |\cos 2\alpha|\).
Squaring: \(4p^2 = k^2 \cos^2 2\alpha\).
Line 2: \(x \sin \alpha + y \cos \alpha = k \sin 2\alpha\).
\[ q = \frac{|-k \sin 2\alpha|}{\sqrt{\sin^2 \alpha + \cos^2 \alpha}} = k |\sin 2\alpha| \]
Squaring: \(q^2 = k^2 \sin^2 2\alpha\).
Adding the two equations:
\[ 4p^2 + q^2 = k^2 (\cos^2 2\alpha + \sin^2 2\alpha) = k^2 \]
Step 3: Final Answer:
\(k^2 = 4p^2 + q^2\), which is option (D).