Quick Summary

O is the vertex of y² = 4x, and P, Q lie on the parabola with OP ⊥ OQ. Parametrise P = (t₁², 2t₁), Q = (t₂², 2t₂); perpendicularity of the slopes 2/t₁ and 2/t₂ forces t₁t₂ = −4, a fixed constant regardless of where P and Q sit individually. Writing the midpoint’s coordinates in terms of s = t₁+t₂ and this fixed product eliminates the parameter entirely and leaves y² = 2(x − 4) — a parabola shifted along the axis, with 4a′ = 2. So the latus rectum of the locus is 2.

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Why This Answer

The key move is recognising that “OP ⊥ OQ from the vertex” is a condition on the PRODUCT t₁t₂, not on t₁ and t₂ individually. Since the vertex O is the origin, the slope of OP is simply 2t₁/t₁² = 2/t₁, and similarly 2/t₂ for OQ. Perpendicular slopes multiply to −1, giving (2/t₁)(2/t₂) = −1, i.e. t₁t₂ = −4 — locked, no matter how P and Q individually move along the curve. That means as P and Q vary (subject to this constraint), the midpoint traces a curve, and eliminating the one free parameter (s = t₁+t₂) from the midpoint’s x and y coordinates gives that curve directly.

Distractor forensics: 1 comes from forgetting the factor of 2 in the shifted latus rectum (using a′ = 1/4 instead of 1/2); 4 is the latus rectum of the original parabola y²=4x, mistakenly carried over to the locus; 8 would come from doubling that original value instead of halving it. The genuine result is that the locus parabola is “half as wide” as the original, reflecting the general pattern for this configuration.

Core Concept

Parametric coordinates and the vertex-chord trick. For y² = 4ax, every point is (at², 2at), and a chord from the vertex O=(0,0) to a point with parameter t has slope 2at/(at²) = 2/t. When two such chords are perpendicular, the product of their parameters is fixed at t₁t₂ = −4a²/a = −4a (for a=1 here, t₁t₂=−4). This is the vertex-analogue of the more famous focal-chord condition t₁t₂=−1, and it appears whenever a problem fixes the angle at the vertex.

Eliminating the parameter for a locus. With s = t₁+t₂ free and p = t₁t₂ = −4 fixed, the midpoint is x = (t₁²+t₂²)/2 = (s²−2p)/2, y = t₁+t₂ = s. Substituting s = y turns the x-equation into a direct relation between x and y — the locus.

Step-by-Step Solution

1. Set up parametric points. For y²=4x (so a=1), write P = (t₁², 2t₁) and Q = (t₂², 2t₂).

2. Impose perpendicularity at O. Slope OP = 2/t₁, slope OQ = 2/t₂. OP⊥OQ ⇒ (2/t₁)(2/t₂) = −1 ⇒ t₁t₂ = −4.

3. Write the midpoint M(x,y) of PQ:

   x = (t1² + t2²)/2, y = (2t1 + 2t2)/2 = t1 + t2

4. Eliminate the parameter. Let s = t₁+t₂ = y. Then t1²+t2² = s² − 2t1t2 = y² − 2(−4) = y² + 8.

   2x = y² + 8 ⇒ y² = 2x − 8 = 2(x − 4)

5. Read off the latus rectum. Comparing y² = 2(x−4) with Y² = 4a′X (X = x−4, Y = y) gives 4a′ = 2, so the latus rectum is 2.

Final Answer

The locus of the midpoint is the parabola y² = 2(x−4), with latus rectum 2. In the exam, the moment you see “chords from the vertex, perpendicular,” reach for the t₁t₂ = −4a shortcut — it converts a locus problem into pure algebra in under a minute.

Key Facts to Remember

• For y²=4ax, a chord from the vertex to parameter t has slope 2/t.
• Two vertex-chords are perpendicular ⇔ t₁t₂ = −4a (contrast with the focal-chord condition t₁t₂ = −1).
• Midpoint locus technique: express x, y of the midpoint via s=t1+t2 and the fixed product, then eliminate s.
• The resulting locus is again a parabola, with latus rectum equal to half that of the original (here 4 → 2).

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Prerequisites

Before practising this question type, make sure the following are in place:

• Parametric coordinates of y²=4ax and the slope of a chord to a parametric point from the vertex.
• Symmetric function identity t1²+t2² = (t1+t2)² − 2t1t2, used to eliminate the free parameter.
• Comparing a shifted parabola to the standard form Y²=4a′X to read off the latus rectum.

Revise these from the Conic Sections chapter, then attempt the related questions below.

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