Question

A curve is traced by a point P(x, y) which moves such that its distance from the point A(-1,1) is three times its distance from the point B(2,-1). Determine the equation of the curve.

Answer 1

`8x^2+8y^2+43-38x+20y=0`

Explanation:

Let
`A\left ( x_1,y_1 \right )` and
`B\left ( x_2,y_2 \right )` be two points then distance AB is equal to
`AB=√(\left ( x_2-x_1 \right )^2+\left ( y_2-y_1 \right )^2)`

Here, a curve is traced by a point
`P(x,y)` which moves such that its distance from the point
`A(-1,1)` is three times its distance from the point
`B(2,-1)` i.e
`AP=3BP`

Using distance formula,

`AP=√((-1-x)^2+(1-y)^2)`

`BP=√((2-x)^2+(-1-y)^2)`

`AP=3BP\\√((-1-x)^2+(1-y)^2)=3√((2-x)^2+(-1-y)^2)`

On squaring both sides, we get

`(-1-x)^2+(1-y)^2=9\left [ (2-x)^2+(-1-y)^2 \right ]\\1+x^2+2x+1+y^2-2y=9\left ( 4 +x^2-4x+1+y^2+2y\right )\\1+x^2+2x+1+y^2-2y=36+9x^2-36x+9+9y^2+18y\\8x^2+8y^2+43-38x+20y=0`

So, equation of curve is
`8x^2+8y^2+43-38x+20y=0`