If P = (x, y),F1 = (3,0), F2 = ( 3, 0) and 16x2 + 25y2 = 400, then PF1 + PF2 equals

Question

If P = (x, y),F1 = (3,0), F2 = ( 3, 0) and 16x2 + 25y2 = 400, then PF1 + PF2 equals (A) 8 (B) 6 (C) 10 (D) 12

Tardigrade Admin · Accepted Answer

Given, 16 x2 + 25y2 = 400 [given] implies (x2/25)+ (y2/16) = 1 Here, a2 = 25, b2 = 16 But b2 = a2 (1- e2) implies 16= 25 (1-e2) implies (16/25)= 1- e2 implies e2 = 1 - (16/25) = (9/25) implies e = (3/5) Now, foci of the ellipse are (± ae, 0) = (± 3 ,0). We have, 3 = a (3/5) implies a = 5 Now, PF1 + PF2 = Major axis = 2a = 2 × 5 = 10

Q. If P=(x,y),F1​=(3,0),F2​=(3,0) and 16x2+25y2=400, then PF1​+PF2​ equals

8

6

10

12

Given, 16x2+25y2=400 [given] ⟹25x2​+16y2​=1 Here, a2=25,b2=16 But b2=a2(1−e2) ⟹16=25(1−e2)⟹2516​=1−e2 ⟹e2=1−2516​=259​⟹e=53​ Now, foci of the ellipse are (±ae,0)=(±3,0). We have, 3=a53​⟹a=5 Now, PF1​+PF2​= Major axis =2a =2×5=10

Q. If $P = (x, y), F_{1} = (3, 0), F_{2} = (3, 0)$ and $16 x^{2} + 25 y^{2} = 400,$ then $P F_{1} + P F_{2}$ equals