Then, the maximum distance of P from the vertex of the parabola x2+6x+y+13=0 is equal to :
(A)8
(B)10
(C)12✓ Correct
(D)9
Answer: (C)
Solution
The circle x2+y2−6x−8y+21=0 has centre C(3,4) and radius 9+16−21=2. The parabola x2+6x+y+13=0, rewritten as (x+3)2=−(y+4), has vertex A(−3,−4). The distance AC=(3+3)2+(4+4)2=36+64=10. The farthest point of the circle from A lies along the ray AC produced, so the maximum distance is AC+r=10+2=12.
JEE Main · 2026 · 05 Apr 2026 · Shift 1 · Single correct option