<p>If a normal to the parabola \(y^2 = 4ax\) makes an angle \(\psi\) with its axis then it will cut the curve again at an angle</p>
<p>A. \(\dfrac{\psi}{2}\)</p>
<p>B. \(\tan^{-1}(2\tan\psi)\)</p>
<p>C. \(\tan^{-1}\!\left(\dfrac{1}{2}\tan\psi\right)\)</p>
<p>D. none of these</p>
Step-by-Step Solution
Key Concept: A normal at parameter t₁ on y² = 4ax has slope -t₁, and when extended to meet the parabola again at parameter t₂, the relationship t₁t₂ = -1 determines the second intersection angle uniquely.
<p><strong>Step 1:</strong> For parabola y² = 4ax, parametric form is (at², 2at). The slope of tangent at parameter t is 1/t, so the normal has slope -t.</p><p><strong>Step 2:</strong> If normal at t₁ makes angle ψ with x-axis, then tan(ψ) = -t₁, so t₁ = -tan(ψ).</p><p><strong>Step 3:</strong> The key property: a normal at t₁ meets the parabola again at t₂ where <strong>t₁t₂ = -1</strong>. Therefore: t₂ = -1/t₁ = -1/(-tan(ψ)) = cot(ψ) = tan(π/2 - ψ).</p><p><strong>Step 4:</strong> The slope of the normal at t₂ is -t₂ = -cot(ψ). The angle θ this normal makes with the x-axis: tan(θ) = -cot(ψ) = -tan(π/2 - ψ) = tan(-(π/2 - ψ)) = tan(ψ - π/2).</p><p><strong>Step 5:</strong> Therefore θ = ψ - π/2, or equivalently, the normal at the second intersection makes an angle <strong>ψ - 90°</strong> or <strong>(ψ - π/2)</strong> with the axis (or equivalently, the acute angle is <strong>90° - ψ</strong>).</p><p>∴ Answer: C</p>
Correct Answer: C