Question

A(-9,0),B(-1,0)are two points.if p is a point such that PA/PB=3/1 find the locus of p​

Answer 1

Step-by-step explanation:

Let the coordinates of P be ( x, y )

$pa = \sqrt{ {(x + 9)}^{2} + {(y - 0)}^{2} } \\ \\ \\ pb = \sqrt{ {(x + 1)}^{2} + {(y - 0)}^{2} } \\ \\ \\ \frac{pa}{pb} = \frac{1}{3} \\ \\ \\ \frac{ \sqrt{ {(x + 9)}^{2} + {y}^{2} } }{ \sqrt{ {(x + 1)}^{2} + {y}^{2} } } = \frac{1}{3} \\ \\ \\ squaring \: both \: sides \\ \\ \\ \frac{ {x}^{2} + 18x + 91 + {y}^{2} }{ {x}^{2} + 2x + 1 + {y}^{2} } = \frac{1}{9} \\ \\ \\ cross \: multiplication \\ \\ \\ 9 {x}^{2} + 162x + 819 + 9{y}^{2} = {x}^{2} + 2x + 1 + {y}^{2} \\ \\ \\ 8 {x}^{2} + 8 {y}^{2} + 160x + 818 = 0$

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Answer 2

Answer:

LET P(x,y) be the point on the locus.

Given points A(-9,0), B(-1,0)

Given Condition PA/PB=3/1

now cross multiplication

PA=3PB

NOW Squaring On Both Sides

PA²=9PB²

P(x,y) A(-9,0) B(-1,0)

(x+9)²+(y-0)²=9(x+1)²+(y-0)²

x²+81+18x+y²=9(x²+1+2x) + y²

x²+81+18x+y²=9x²+9+18x+y²

canceling both sides y² and 18x

now 9x²+9-x²-81= 8x²-72

common 8 (x²-9)=0

therefor The locus is X²-9=0

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