Straight Lines
Distance from Point to Line
Grade 11
Question:
<p>If <i>p</i> and <i>p</i>′ be perpendiculars from the origin upon the straight lines \(x \sec \theta + y \csc \theta = a\) and \(x \cos \theta - y \sin \theta = a \cos 2\theta\) respectively, then the value of the expression \(4p^2 + p'^2\) is</p>
<p>(a) \(a^2\)</p>
<p>(b) \(3a^2\)</p>
<p>(c) \(2a^2\)</p>
<p>(d) \(4a^2\)</p>
Step-by-Step Solution
Key Concept: Use the perpendicular distance formula from origin to a line and apply trigonometric identities to simplify the sum.
<p><strong>Step 1:</strong> Find perpendicular distance <i>p</i> from origin to the line $x \sec \theta + y \csc \theta = a$:</p><p>$$p = \frac{a}{\sqrt{\sec^2 \theta + \csc^2 \theta}} = \frac{a \sin^2 \theta}{1}$$</p><p>$\therefore 2p = a \sin^2 \theta$ ... (i)</p><p><strong>Step 2:</strong> Find perpendicular distance <i>p</i>′ from origin to the line $x \cos \theta - y \sin \theta = a \cos 2\theta$:</p><p>$$p' = \frac{a \cos 2\theta}{\sqrt{\cos^2 \theta + \sin^2 \theta}} = a \cos 2\theta$$</p><p>... (ii)</p><p><strong>Step 3:</strong> Square and add equations (i) and (ii):</p><p>$$4p^2 + p'^2 = (a \sin^2 \theta)^2 + (a \cos 2\theta)^2 = a^2(\sin^4 \theta + \cos^2 2\theta)$$</p><p>Using $\cos 2\theta = 1 - 2\sin^2 \theta$, we get:</p><p>$$4p^2 + p'^2 = a^2$$</p><p>∴ Answer is (a).</p>
Correct Answer: a