JEE Main202622 January 2026Evening ShiftMathematicsEllipseActual
Let S and S ^ be the foci of the ellipse x² 25 + y² 9 =1 and P ( , ) be a point on the ellipse in the first quadrant. If ( SP )²+ ( S ^ P )²- SP S ^ P =37 , then ²+ ² is equal to :
Options
- A15
- B11
- C17
- D13
Correct answer
D. 13
Step-by-step solution
For the ellipse x^2 25 + y^2 9 = 1 , we have a = 5 , b = 3 , c = 4 . So foci are at S( 4, 0) . For point P on the ellipse: SP + S'P = 10 . Given (SP)^2 + (S'P)^2 - SP S'P = 37 Let r₁ = SP and r₂ = S'P . From (r₁ + r₂)^2 = 100 , we get r₁^2 + r₂^2 = 100 - 2r₁r₂ . Substituting into the given equation: 100 - 3r₁r₂ = 37 , so r₁r₂ = 21 . Thus r₁ = 3, r₂ = 7 . From ( + 4)^2 + ^2 = 9 and ( - 4)^2 + ^2 = 49 , subtracting gives 16 = -40 , so = - 5 2 . From the ellipse equation: ^2 = 27 4 . Therefore ^2 + ^2 = 25 4 + 27 4 =