JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Parabola

Let

A, B

and C be the vertices of a variable right angled triangle inscribed in the parabola

y^{2} = 16 x

. Let the vertex B containing the right angle be

(4, 8)

and the locus of the centroid of

△ ABC

be a conic

C_{0}

. Then three times the length of latus rectum of

C_{0}

is ____.

Answer: 16

Solution

Parametrise

y^{2} = 16 x

(4 t^{2}, 8 t)

. The vertex

B = (4, 8)

corresponds to

t = 1

; let

A = (4 t_{1}^{2}, 8 t_{1})

C = (4 t_{2}^{2}, 8 t_{2})

. The slope of

B A

\frac{8 t _{1} - 8}{4 t _{1}^{2} - 4} = \frac{2}{t _{1} + 1}

and similarly slope of

B C

\frac{2}{t _{2} + 1}

. The right angle at

B

gives

\frac{4}{( t _{1} + 1 ) ( t _{2} + 1 )} = - 1

, so

(t_{1} + 1) (t_{2} + 1) = - 4

, i.e.

t_{1} t_{2} + (t_{1} + t_{2}) = - 5

. Writing

s = t_{1} + t_{2}

, the centroid

(h, k)

satisfies

3 k = 8 + 8 s

and

3 h = 4 + 4 (t_{1}^{2} + t_{2}^{2}) = 4 + 4 (s^{2} - 2 t_{1} t_{2})

. Using

t_{1} t_{2} = - 5 - s

3 h = 4 + 4 (s^{2} + 2 s + 10)

. Substituting

s = \frac{3 k - 8}{8}

gives

3 h = \frac{9 k ^{2}}{16} + 40

, so the locus is

h = \frac{3}{16} k^{2} + \frac{40}{3}

, equivalently

y^{2} = \frac{16}{3} (x - \frac{40}{3})

. Its latus rectum is

\frac{16}{3}

, and three times this is

16

JEE Main · 2026 · 04 Apr 2026 · Shift 2 · Integer answer