Equation of hyperbola is
a2(x−6)2−4−a2(y−2)2=1
⇒(4−a2)(x−6)2−a2(y−2)2=a2(4−a2)
comparing with 3x2−y2−αx+βy+γ=0, we get a2=1 and α=36,β=4 and γ=101
∴α+β+γ=141
JEE Main 2025 — Mathematics Coordinate Geometry
2025integereasy
If the equation of the hyperbola with foci (4,2) and (8,2) is 3x2−y2−αx+βy+γ=0, then α+β+γ is equal to _____.
Official previous-year question
Held on 3 Apr 2025 · Verified 6 Jul 2026.
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