0:00:00.430,0:00:04.430
Jom kita cuba cth lain bagi Lokus.

0:00:05.108,0:00:10.488
Diberi titik tetap A, B dan [br]titik bergerak P;

0:00:15.718,0:00:25.468
PA : PB = 2 : 1

0:00:26.732,0:00:30.612
Jadi apakah cara utk dapatkan [br]persamaan Lokus ini?

0:00:31.005,0:00:33.525
Mudah saja.

0:00:33.707,0:00:36.787
Tukarkan ke dlm bentuk pecahan.

0:00:36.916,0:00:41.296
PA/PB = 2/1

0:00:46.721,0:00:48.631
Darab silang persamaan ini.

0:00:48.888,0:00:49.728
Kita akan dapat:

0:00:49.969,0:00:54.769
PA = 2PB

0:01:00.758,0:01:04.548
Gunakan formula jarak.

0:01:04.745,0:01:13.275
PA: [br]√ [ (x - 3)2 + (y-4)2]

0:01:14.200,0:01:16.260
= 2PB,

0:01:20.538,0:01:32.788
PB:[br]√ [ (x - 1)2 + (y - (-6))2]

0:01:37.627,0:01:43.507
Ingat, kuasa-dua kan persamaan [br]utk hapuskan √

0:01:43.584,0:01:44.764
Hasilnya,

0:01:44.939,0:02:08.089
(x - 3)2 + (y - 4)2 = 4 [ (x-1)2 + (y+6)2]

0:02:08.266,0:02:09.226
Kemudian, kembangkan persamaan ini.

0:02:09.393,0:02:33.523
x2 - 6x + 9 + y2 - 8y + 16 = [br]4 [ x2 - 2x + 1 + y2 + 12y + 36]

0:02:33.681,0:02:34.711
Kembangkan lagi PB:

0:02:35.018,0:02:48.508
4x2 - 8x + 4 + 4y2 + 48y + 144,

0:02:48.718,0:02:58.278
Dan PA:[br]x2 - 6x + y2 - 8y + 25

0:03:01.487,0:03:05.487
Kumpulkan semua nilai di sebelah kanan (PB).

0:03:07.712,0:04:03.242
3x2 - 2x + 3y2 + 56y + 4 + 144 - 25 = 0

0:04:03.640,0:04:05.200
Susun dan ringkaskan.

0:04:05.474,0:04:18.264
0 = 3x2 + 3y2 - 2x + 56y + 123

0:04:18.738,0:04:22.738
Inilah persamaan Lokus yg kita cari.