Question

A(-9,0), B(-1,0)
are two points. If P is a
point such that PA: PB = 3:1, then the locus
of Pis​

Answer 1

Answer:

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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