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5.1 · Q3

Q.If A(2, 0) and B(0, 3) are two points, find the equation of the locus of point P such that AP = 2BP.

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Let P(x,y)P(x,y) satisfy AP=2BPAP=2BP, so AP2=4BP2AP^2=4BP^2:

(x−2)2+y2=4[x2+(y−3)2].(x-2)^2+y^2=4\left[x^2+(y-3)^2\right].

Expand the left: x2−4x+4+y2x^2-4x+4+y^2. Expand the right: 4x2+4y2−24y+364x^2+4y^2-24y+36.

So x2−4x+4+y2=4x2+4y2−24y+36x^2-4x+4+y^2=4x^2+4y^2-24y+36.

Bring everything to one side: x2−4x+4+y2−4x2−4y2+24y−36=0⇒−3x2−3y2−4x+24y−32=0x^2-4x+4+y^2-4x^2-4y^2+24y-36=0 \Rightarrow -3x^2-3y^2-4x+24y-32=0.

Multiply by −1-1: 3x2+3y2+4x−24y+32=03x^2+3y^2+4x-24y+32=0.

✓Final answer

3x2+3y2+4x−24y+32=03x^2 + 3y^2 + 4x - 24y + 32 = 0

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