ParabolahardPYQ · JEE Main · 4 Apr 2026 · Shift 2 (Afternoon)Free

Locus of Centroid, Right Triangle in Parabola: 3·LR = 16 | JEE 2026

JEE Maths question with a full step-by-step solution.

Question
Let A,BA,B and CC be the vertices of a variable right-angled triangle inscribed in the parabola y2=16xy^2=16x. Let the vertex BB containing the right angle be (4,8)(4,8) and the locus of the centroid of △ABC\triangle ABC be a conic C0C_0. Then three times the length of the latus rectum of C0C_0 is
Solution
Answer: 16 (± 0.01)
Step 1: A(4t12,8t1)A(4t_1^2,8t_1), C(4t22,8t2)C(4t_2^2,8t_2), B(4,8)B(4,8) (t=1t=1). mAB=2t1+1m_{AB}=\dfrac{2}{t_1+1}, mBC=2t2+1m_{BC}=\dfrac{2}{t_2+1}. Right angle at BB:
2t1+1⋅2t2+1=−1⇒4=−(t1+1)(t2+1)\frac{2}{t_1+1}\cdot\frac{2}{t_2+1}=-1\Rightarrow 4=-(t_1+1)(t_2+1)
⇒t1t2+t1+t2+1=−4⇒t1+t2+t1t2=−5.(1)\Rightarrow t_1t_2+t_1+t_2+1=-4\Rightarrow t_1+t_2+t_1t_2=-5.\quad(1)
 Step 2: Centroid (h,k)(h,k), with s=t1+t2, p=t1t2s=t_1+t_2,\ p=t_1t_2:
3h=4+4t12+4t22=4+4(s2−2p),3k=8+8s.3h=4+4t_1^2+4t_2^2=4+4(s^2-2p),\qquad 3k=8+8s.
 From (1): p=−5−sp=-5-s. Step 3: s=3k−88s=\dfrac{3k-8}{8}. Sub p=−5−sp=-5-s:
3h=4+4s2−8(−5−s)=4s2+8s+44.3h=4+4s^2-8(-5-s)=4s^2+8s+44.
3h=4(3k−88)2+8⋅3k−88+44=(3k−8)216+(3k−8)+44.3h=4\left(\frac{3k-8}{8}\right)^2+8\cdot\frac{3k-8}{8}+44=\frac{(3k-8)^2}{16}+(3k-8)+44.
 (3k−8)2=9k2−48k+64(3k-8)^2=9k^2-48k+64:
3h=9k2−48k+6416+3k−8+44=9k216−3k+4+3k+36=9k216+40.3h=\frac{9k^2-48k+64}{16}+3k-8+44=\frac{9k^2}{16}-3k+4+3k+36=\frac{9k^2}{16}+40.
 ⇒h=3k216+403\Rightarrow h=\dfrac{3k^2}{16}+\dfrac{40}{3}. ∴C0:\therefore C_0:
x=948y2+403.x=\frac{9}{48}y^2+\frac{40}{3}.
 Step 4:
y2=489(x−403)⇒4a=489.y^2=\frac{48}{9}\left(x-\frac{40}{3}\right)\Rightarrow 4a=\frac{48}{9}.
∴3×489=1449=16.\therefore 3\times\frac{48}{9}=\frac{144}{9}=16.
 Correct answer: 16
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