Parabola
Chord of Parabola
Grade 11
Question:
<p>Let chord be \(y = m\left(x - \dfrac{3}{2}\right)\) or \(mx - y - \dfrac{3m}{2} = 0\) for the parabola \(y^2 = 8x\). The chord is at distance \(\dfrac{\sqrt{5}}{2}\) from the origin. Find the slope \(m\).</p>
<p>\(m = \pm 1\)</p>
<p>\(m = \pm \dfrac{\sqrt{5}}{2}\)</p>
<p>\(m = \pm 2\)</p>
<p>\(m = \pm \dfrac{1}{2}\)</p>
Step-by-Step Solution
Key Concept: Use the point-to-line distance formula d = |ax₀ + by₀ + c|/√(a² + b²) with origin (0,0) as the point. The chord equation mx - y - 3m/2 = 0 gives distance √5/2, which yields a quadratic in m after algebraic manipulation.
<p><strong>Step 1:</strong> Apply the distance formula from origin (0,0) to line mx - y - 3m/2 = 0:</p><p>d = |m(0) - 0 - 3m/2|/√(m² + 1) = |3m/2|/√(m² + 1)</p><p><strong>Step 2:</strong> Set distance equal to √5/2:</p><p>|3m/2|/√(m² + 1) = √5/2</p><p><strong>Step 3:</strong> Square both sides:</p><p>9m²/4(m² + 1) = 5/4</p><p>9m² = 5(m² + 1)</p><p>9m² = 5m² + 5</p><p>4m² = 5</p><p>m² = 5/4</p><p><strong>Step 4:</strong> Solve for m:</p><p>m = ±√(5/4) = ±√5/2</p><p>∴ Answer: B</p>
Correct Answer: B