Question

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

Answer 1

Answer:  The required equation of the locus of point P is $3x^2+3y^2+4x-24y+32=0.$

Step-by-step explanation:  Given that A(2,0) and B(0,3) are two points.

We are to find the locus of point P such that AP=2BP.

We will be using the following distance formula :

Distance formula :  The distance between two points X(a, b) and Y(c, d) is given by

$XY=\sqrt{(c-a)^2+(d-b)^2}.$

Let P(x, y) be the given point.

According to the given information, we have

$AP=2BP\\\\\Rightarrow \sqrt{(x-2)^2+(y-0)^2}=2\sqrt{(x-0)^2+(y-3)^2}\\\\\Rightarrow (x-2)^2+y^2=4(x^2+(y-3)^2)~~~~~~~~[\textup{Squaring both sides}]\\\\\Rightarrow x^2-4x+4+y^2=4x^2+4y^2-24y+36\\\\\Rightarrow 3x^2+3y^2+4x-24y+32=0.$

Thus, the required equation of the locus of point P is $3x^2+3y^2+4x-24y+32=0.$

Answer 2

Answer:

$\red { Equation \: of \: the \:locus \:of \:point \:P}$

$\green { \: 3x^{2} + 3y^{2} + 4x - 24y + 32 = 0}$

Step-by-step explanation:

$Given \: two \:points \:A(2,0) \:and \:B(0,3)$

$Let \: P(x,y) \: is \: a \: point \: such \:that \\AP = 2BP$

$\underline {\pink { By \:distance \: formula : }}$

$Distance \: between \: two \: points \\(x_{1},y_{1}) \:and \:(x_{2},y_{2}) \:is$

$\orange {\sqrt{ (x_{2}-x_{1})^{2} + (y_{2}-y_{1})^{2} }}$

$AP = 4BP$

$\implies AP^{2} = 4BP^{2}$

$\implies (x-2)^{2} + y^{2} = 4[ x^{2} + (y-3)^{2} ]$

$\implies x^{2} - 4x + 4 = 4( x^{2} + y^{2} -6y+9 )$

$\implies x^{2} - 4x + 4 = 4x^{2} + 4y^{2} -24y+36$

$\implies 0 =4x^{2} + 4y^{2} -24y+36 -x^{2} +4x - 4$

$\implies 3x^{2} + 3y^{2} + 4x - 24y + 32 = 0$

Therefore.,

$\red { Equation \: of \: the \:locus \:of \:point \:P}$

$\green { \: 3x^{2} + 3y^{2} + 4x - 24y + 32 = 0}$

•••♪