Parabola: y=3a3x2+2a2x−2a Vertex: (α,β) α=2a3/3−a2/2=−4a3,β=43a3−(4a4+4⋅3a3⋅2a)=−34a3−(41+38)a4=−12354a×3=−1635aαβ=−4a3(−1635)a=64105.
The locus of the vertices of the family of parabolas y= a^3 x^2 3+ a^2 x 2-2 a is
JEE Main 2006 — Mathematics Coordinate Geometry
The locus of the vertices of the family of parabolas y=3a3x2+2a2x−2a is
Held on 30 Apr 2006 · Verified 6 Jul 2026.
Options
- A
xy=64105
- B
xy=43
- C
xy=1635
- D
xy=10564
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