Question

Find the equation of the locus of P, A=(2,3) B=(2,-3) and PA+PB =8​

Answer 1

Step-by-step explanation:

Given A = (2, 3) B = (2, –3) Let P(x, y) be any point on the locus. Given geometric condition is PA + PB = 8 ⇒ PA = 8 – PB ⇒ PA2 = (8 – PB)2 ⇒ PA2 = 64 + PB2 – 16PB ⇒ (x – 2)2+(y – 3)2 = 64 + (x – 2)2 + (y + 3)2 - 16√((x - 2)2 + (y + 3)2) ⇒ y2 – 6y + 9 – y2 – 6y – 9 – 64 = – 16√(x2 - 4x + 4 + y2 + 6y + 9) ⇒ (–12y – 64) = –16√(x2 + y2 - 4x + 6y + 13) ⇒ (3y + 16) = 4√(x2 + y2 - 4x + 6y + 13) S.O.B.S ⇒ (3y + 16)2 = 16 (x2 + y2 – 4x + 6y + 13) ⇒ 9y2 + 256 + 96y = 16x2 + 16y2 – 64x + 96y + 208 ⇒ 16x2 + 7y2 – 64x – 48 = 0 ∴ The locus of P is 16x2 + 7y2 – 64x – 48 = 0Read more on Sarthaks.com - https://www.sarthaks.com/533033/find-the-equation-of-the-locus-of-p-if-a-2-3-b-2-3-and-pa-pb-8

Answer 2

Step-by-step explanation:

this is my answer: 16x²-64x+7y²=48.