Ellipse
Normal to Ellipse
Grade 11
Question:
<p>If \(\beta\) is one of the angles between the normal to the ellipse, \(x^2 + 3y^2 = 9\) at the points \((3\cos\theta, \sqrt{3}\sin\theta)\) and \((-3\sin\theta, \sqrt{3}\cos\theta)\); \(\theta \in (0, \pi/2)\); then \(\dfrac{2\cot\beta}{\sin 2\theta}\) is equal to</p>
<p>\(\dfrac{2}{\sqrt{3}}\)</p>
<p>\(\dfrac{1}{\sqrt{3}}\)</p>
<p>\(\sqrt{2}\)</p>
<p>\(\dfrac{\sqrt{3}}{4}\)</p>
Step-by-Step Solution
Key Concept: The normal to an ellipse at point (x₀, y₀) has slope found by implicit differentiation. Two normals at different points have an angle β between them; use the tangent subtraction formula: tan(β) = |(m₁ - m₂)/(1 + m₁m₂)| where m₁, m₂ are the slopes of the normals.
<p><strong>Step 1: Find the normal at P₁ = (3cos θ, √3 sin θ)</strong></p><p>For ellipse x² + 3y² = 9, differentiating implicitly: 2x + 6y(dy/dx) = 0 ⟹ dy/dx = -x/(3y)</p><p>At P₁: slope of tangent = -3cos θ/(3√3 sin θ) = -cos θ/(√3 sin θ)</p><p>Slope of normal m₁ = √3 sin θ/cos θ = √3 tan θ</p><p><strong>Step 2: Find the normal at P₂ = (-3sin θ, √3 cos θ)</strong></p><p>At P₂: slope of tangent = -(-3sin θ)/(3√3 cos θ) = sin θ/(√3 cos θ)</p><p>Slope of normal m₂ = -√3 cos θ/sin θ = -√3 cot θ</p><p><strong>Step 3: Apply angle formula between two lines</strong></p><p>tan β = |m₁ - m₂|/|1 + m₁m₂| = |√3 tan θ - (-√3 cot θ)|/|1 + (√3 tan θ)(-√3 cot θ)|</p><p>= |√3(tan θ + cot θ)|/|1 - 3| = √3(tan θ + cot θ)/2</p><p><strong>Step 4: Simplify using sin 2θ</strong></p><p>tan θ + cot θ = sin θ/cos θ + cos θ/sin θ = (sin² θ + cos² θ)/(sin θ cos θ) = 1/(sin θ cos θ) = 2/sin 2θ</p><p>Therefore: tan β = √3 · (2/sin 2θ)/2 = √3/sin 2θ</p><p>cot β = sin 2θ/√3</p><p><strong>Step 5: Calculate final expression</strong></p><p>2cot β/sin 2θ = 2 · (sin 2θ/√3)/sin 2θ = 2/√3 = 2√3/3</p><p>∴ Answer: <strong>2/√3 or 2√3/3</strong></p>
Correct Answer: B