Parabola
Normal to Parabola
Grade 11
Question:
<p>The equation of normal at a point \('t'\), i.e., at \((at_1^2, 2at_1)\) on the parabola \(y^2 = 4bx\) is given by \(y = -t_1 x + 2at_1 + at_1^3\). If this normal intersects the parabola again at point \('t'\), then \(t\) equals:</p>
<p>(A) \(t = \dfrac{-1 + (1 + t_1^2)}{t_1}\)</p>
<p>(B) \(t = \dfrac{-1 \pm (1 + t_1^2)}{t_1}\)</p>
<p>(C) \(t = \dfrac{1 \pm (1 + t_1^2)}{t_1}\)</p>
<p>(D) \(t = \dfrac{-1 - (1 + t_1^2)}{t_1}\)</p>
Step-by-Step Solution
Key Concept: A normal at parameter t₁ on parabola y²=4ax intersects the parabola at another point with parameter t satisfying t₁·t = -2. This comes from the condition that if a line intersects the parabola at parameters t₁ and t, their product relation depends on the line's slope and intercept properties.
<p><strong>Step 1:</strong> The normal at point (at₁², 2at₁) on parabola y² = 4ax has equation y = -t₁x + 2at₁ + at₁³</p><p><strong>Step 2:</strong> For a normal at parameter t₁, if it intersects the parabola again at parameter t, use the focal chord property: the product of parameters for a normal is t₁·t = -2</p><p><strong>Step 3:</strong> However, the standard result is that if a normal at t₁ meets the parabola again at t, then: t = -(t₁ + 2/t₁)</p><p><strong>Step 4:</strong> Alternatively, applying the parametric intersection condition directly: when the normal line intersects y² = 4ax, substituting and factoring gives parameters whose product follows t·t₁ = -2, yielding <strong>t = -t₁ - 2/t₁</strong></p><p>∴ Answer: <strong>B</strong> (The specific form depends on options provided; typically t = -(t₁ + 2/t₁) or t = -2/t₁ - t₁)</p>
Correct Answer: B