Let S and S^ ' be the foci of the ellipse x^225+ y^29=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 :

Question

Accepted Answer

Correct answer: D. (d) For the ellipse x^225+ y^29=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 r_1= SP and r_2=S^ 'P. From (r_1+r_2)^2=100, we get r_1^2+r_2^2=100-2r_1r_2. Substituting into the given equation: 100-3r_1r_2=37, so r_1r_2=21. Thus r_1=3 ,r_2=7. From ( +4 )^2+ ^2=9 and ( -4 )^2+ ^2=49, subtracting gives 16 =-40, so =- 52. From the ellipse equation: ^2= 274. Therefore ^2+ ^2= 254+ 274=13.

$Let S and S^{’} be the foci of the ellipse \frac{x ^{2}}{25} + \frac{y ^{2}}{9} = 1 and P (α, β) be a point on the ellipse in the first quadrant. If (SP)^{2} + (S^{’} P)^{2} - SP \cdot S^{’} P = 37, then α^{2} + β^{2} is equal to :$

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