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Jom kita cuba cth lain bagi Lokus.

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Diberi titik tetap A, B dan 
titik bergerak P;

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PA : PB = 2 : 1

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Jadi apakah cara utk dapatkan 
persamaan Lokus ini?

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Mudah saja.

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Tukarkan ke dlm bentuk pecahan.

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PA/PB = 2/1

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Darab silang persamaan ini.

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Kita akan dapat:

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PA = 2PB

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Gunakan formula jarak.

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PA: 
√ [ (x - 3)2 + (y-4)2]

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= 2PB,

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PB:
√ [ (x - 1)2 + (y - (-6))2]

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Ingat, kuasa-dua kan persamaan 
utk hapuskan √

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Hasilnya,

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(x - 3)2 + (y - 4)2 = 4 [ (x-1)2 + (y+6)2]

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Kemudian, kembangkan persamaan ini.

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x2 - 6x + 9 + y2 - 8y + 16 = 
4 [ x2 - 2x + 1 + y2 + 12y + 36]

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Kembangkan lagi PB:

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4x2 - 8x + 4 + 4y2 + 48y + 144,

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Dan PA:
x2 - 6x + y2 - 8y + 25

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Kumpulkan semua nilai di sebelah kanan (PB).

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3x2 - 2x + 3y2 + 56y + 4 + 144 - 25 = 0

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Susun dan ringkaskan.

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0 = 3x2 + 3y2 - 2x + 56y + 123

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Inilah persamaan Lokus yg kita cari.