f1≡(−ae,0)≡P(−3,0)⇒ae=3

tan45∘=2aeb2/a2ae=ab2b2=6a Also a2e2=a2+b29=a2+6aa2+6a−9=0a=−3±32=−3(1±2)∴a2b2=a2⋅6a=6a3=6(1352−189)α=810 and β=1134∴α+β=1944
JEE Main 2025 — Mathematics Coordinate Geometry
Consider the hyperbola a2x2−b2y2=1 having one of its focus at P(−3,0). If the latus ractum through its other focus subtends a right angle at P and a2b2=α2−β,α,β∈N.
Held on 7 Apr 2025 · Verified 6 Jul 2026.
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