Show that the sum of the focal distances of any point on an ellipse is equal to the length of the major axis and also prove that the locus of a point which moves so that the sum of its distances from 3 , 0 , and - 3 , 0 is 9 , is x 2 81 4 + y 2 45 4 = 1

To Prove: The sum of the focal distances of any point on an ellipse is equal to the length of the major axis. Consider the ellipse x 2 a 2 + y 2 b 2 = 1 . . . . i , where a > b , and whose diagram is given below: The equation of directrices of this ellipse is x = ± a e , where e is the eccentricity of this ellipse. Here, we need to prove that F 1 P + F 2 P = 2 a We know that eccentricity of an ellipse can be taken as the ratio of its distance from the focus and the distance from the directrix. Thus, F 1 P P M = e , and F 2 P P M ' = e ⇒ F 1 P = e P M , and F 2 P = e P M ' So, F 1 P + F 2 P = e P M + e P M ' = e P M + P M ' = e M M ' = e 2 a e [Since the distance between the directrices is 2 a e ] = 2 a . . . . . i Hence, Proved. Now, according to the question, we are given that F 1 = 3 , 0 , F 2 = - 3 , 0 , and 2 a = 9 Thus, we have a = 9 2 ⇒ a 2 = 81 4 . . . . i i Since the coordinates of foci are ± a e , 0 ⇒ a e = 3 Now, b 2 = a 2 - a 2 e 2 = 9 2 2 - 3 2 = 81 4 - 9 = 45 4 . . . . . i i i Hence, from i , i i , and i i i The required equation of an ellipse is x 2 81 4 + y 2 45 4 = 1 .