Question

find the equation of a point "P" if the distance of "P" from A(3,0) is twice the distance of "P" from B(-3,0).

Answer 1

It is not called equation of P. it is called locus of P (x,y)
AP² = 4 PB²
(x-3)² + (y-0)²  = 4 (x+3)² + 4 (y-0)²
x² -6x +9 +y² = 4 x² + 24x + 36 + 4y²
Take all terms to RHS
3x² + 30x +3y² + 27 = 0  =>  x²+10x + y²+9 = 0

Answer 2

Answer:

The locus of the point P(x, y) is $x^{2}+10x+y^{2}+9=0$

Step-by-step explanation:

Given a point A(3, 0) and B(-3, 0)

Given the distance of point P from A is equal to the twice of the distance of point from B.

The distance between two points $A(x_1, y_1)$ and $B(x_2, y_2)$ is given by $d(A,B)=\sqrt{(x_2-x_1)^{2}+ (y_2-y_1)^{2}}$

Given a condition that should give the locus of P.

A locus is a curve formed by all the points satisfying a condition, relating the coordinates.

So, let us assume a point $P(x,~y)$ on the locus.

Then from the given condition, we get

AP = 2BP

Squaring on both sides,  

$(AP)^2 = (2BP)^2$

$\implies (AP)^2 = 4 (PB)^2$

Using the distance formula and substituting the values,

$\implies \big(\sqrt{(x-3)^{2}+ (y-0)^{2}} \big)^2 = 4\big (\sqrt{(x-(-3))^{2}+ (y-0)^{2}}\big)^2$

$\implies (x-3)^{2}+ (y-0)^{2}} = 4({(x-(-3))^{2}+ (y-0)^{2}})$

$\implies x^{2}+9-6x+y^{2}} = 4x^{2}+36+24x+4y^{2}}$

$\implies x^{2}+9-6x+y^{2}} - 4x^{2}-36-24x-4y^{2}}=0$

$\implies -30x- 3x^{2}-27-3y^{2}=0$

$\implies x^{2}+10x+y^{2}+9=0$  

Therefore, the locus of the point P(x, y) is $x^{2}+10x+y^{2}+9=0$