What is the locus of midpoint of PQ when OP⊥OQ on parabola y²=4x?

y²=2(x−4). Using parametric points (t₁²,2t₁) and (t₂²,2t₂) with t₁t₂=−4, the midpoint satisfies k²=2(h−4).

What is the latus rectum of y²=2(x-4)?

Comparing with y²=4a(x−h₀): 4a=2, so a=1/2. Latus rectum = 4a = 2.

Parabola y²=4x, vertex O, chords OP and OQ perpendicular — locus of midpoint of PQ is conic C — latus rectum

JEERANKERS

MathParabola

QMCQConic Sections

Let $O$ be the vertex of the parabola $y^2=4x$ and its chords $OP$ and $OQ$ are perpendicular to each other. If the locus of the mid-point of $PQ$ is a conic $C$, then the length of its latus rectum is:

A) $1$ B) $2$ C) $4$ D) $8$

✅ Correct Answer

B) 2

✓

Solution

Parametric points on parabola

Any point on $y^2=4x$ is $(t^2,2t)$ for parameter $t$.

Let $P=(t_1^2,2t_1)$ and $Q=(t_2^2,2t_2)$. The vertex is $O=(0,0)$.

Apply perpendicularity condition for OP ⊥ OQ

$\text{slope}(OP)=\dfrac{2t_1}{t_1^2}=\dfrac{2}{t_1},\qquad\text{slope}(OQ)=\dfrac{2}{t_2}$

$OP\perp OQ\ \Rightarrow\ \dfrac{2}{t_1}\cdot\dfrac{2}{t_2}=-1\ \Rightarrow\ t_1t_2=-4$

Find locus of midpoint of PQ

Let midpoint $M=(h,k)$:

$h=\dfrac{t_1^2+t_2^2}{2},\qquad k=\dfrac{2t_1+2t_2}{2}=t_1+t_2$

$t_1^2+t_2^2=(t_1+t_2)^2-2t_1t_2=k^2-2(-4)=k^2+8$

$\therefore\quad 2h=k^2+8\ \Rightarrow\ k^2=2(h-4)$

Replacing $(h,k)$ with $(x,y)$: the locus is $y^2=2(x-4)$.

Find the latus rectum

$y^2=2(x-4)$ is a parabola with vertex $(4,0)$. Comparing with $y^2=4a(x-h)$:

$4a=2\ \Rightarrow\ a=\dfrac{1}{2}$

Latus rectum $=4a=\boxed{2}$

📘 Locus on Parabola Using Parametric Form

For $y^2=4ax$, parametric point is $(at^2,2at)$. The slope of a chord through vertex from parameter $t$ is $2/t$. For two perpendicular chords $OP$ and $OQ$: $(2/t_1)(2/t_2)=-1$ gives $t_1t_2=-4a$. Midpoint locus is found by expressing $h=\frac{t_1^2+t_2^2}{2}$ and $k=t_1+t_2$ and using $k^2=t_1^2+t_2^2+2t_1t_2$ identity.

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Important Concepts

Parametric Form of ParabolaFor $y^2=4x$ (here $a=1$): point is $(t^2,2t)$. Slope of line from origin to $(t^2,2t)$ is $2t/t^2=2/t$. This parametric slope is crucial for perpendicularity conditions.

Perpendicular Chord ConditionTwo lines with slopes $m_1$ and $m_2$ are perpendicular iff $m_1m_2=-1$. For chords through vertex: $(2/t_1)(2/t_2)=-1\Rightarrow t_1t_2=-4$. (For general parabola $y^2=4ax$: $t_1t_2=-4a^2/1=-4$.)

Midpoint Locus TechniqueExpress $h$ and $k$ in terms of $t_1+t_2$ and $t_1t_2$. Use $t_1^2+t_2^2=(t_1+t_2)^2-2t_1t_2=k^2+8=2h$. This gives the Cartesian locus equation directly.

Latus Rectum FormulaFor $y^2=4a(x-h_0)$: latus rectum $=4a$. Here $y^2=2(x-4)$ means $4a=2$, $a=1/2$, latus rectum $=2$.

FAQs

What is the parametric form of y²=4x?

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$(t^2,2t)$ for any real $t$. Substituting: $(2t)^2=4t^2$ ✓. The slope from origin to this point is $2t/t^2=2/t$.

Why does t₁t₂=−4?

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Perpendicular slopes: $(2/t_1)(2/t_2)=-1\Rightarrow 4/(t_1t_2)=-1\Rightarrow t_1t_2=-4$.

How do you find the vertex of the locus conic?

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$y^2=2(x-4)$ has vertex $(4,0)$ — the point where the conic crosses its own axis of symmetry.

What type of conic is y²=2(x-4)?

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A parabola opening rightward with vertex at $(4,0)$, axis along $y=0$, focus at $(4+1/2,0)=(4.5,0)$.

Is this from JEE Main 2026?

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Yes, this question appeared in JEE Main 2026.

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