Question

If the distance from P to the points (2, 3) and (2, -3) are in the ratio 2 : 3, then find the equation of the locus of P.

Answer 1

5x^2+5y^2-20x-78y+65=0 is required locus at p(x,y)

Answer 2

Answer:

Equation of the locus of point P is $5x^2+5y^2-20x-78y+65=0$

Step-by-step explanation:

Given: Coordinates of the points say Q( 2 , 3 ) and  R( 2 , -3 )

           Distance of Point P to given point is in ratio 2 : 3

To find: Equation of locus of point P.

Let ( x , y ) be the coordinate of the point P.

Distance between two points is given by $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

According to the Question,

$\frac{PQ}{PR}=\frac{2}{3}$

$\frac{\sqrt{(x-2)^2+(y-3)^2}}{\sqrt{(x-2)^2+(y+3)^2}}=\frac{2}{3}$

$\frac{(x-2)^2+(y-3)^2}{(x-2)^2+(y+3)^2}=\frac{4}{9}$

$9(x^2+2^2-4x+y^2+3^2-6y)=4(x^2+2^2-4x+y^2+3^2+6x)$

$9x^2+36-36x+9y^2+81-54y=4x^2+16-16x+4y^2+36+24y$

$9x^2-4x^2+36-36-36x+16x+9y^2-4y^2+81-16-54y-24y=0$

$5x^2+5y^2-20x-78y+65=0$

Therefore, Equation of the locus of point P is $5x^2+5y^2-20x-78y+65=0$