Question

find the locus of the point whose distance from origin is to its distance from (-2,-3) is as 5:7​

Answer 1

The required locus of the point (x, y) is $24x^2+24y^2-100x-150y-325=0.$

Step-by-step explanation:  Let the required point be represented by (x, y).

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

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

According to the given information, we have

$\dfrac{\textup{distance between (x,y) and (0,0)}}{\textup{distance between (x,y) and (-2,-3)}}=\dfrac{5}{7}\\\\\\\Rightarrow \dfrac{\sqrt{(x-0)^2+(y-0)^2}}{\sqrt{(x+2)^2+(y+3)^2}}=\dfrac{5}{7}\\\\\\\Rightarrow \dfrac{x^2+y^2}{x^2+4x+4+y^2+6y+9}=\dfrac{25}{49}~~~~~~~[\textup{Squaring both sides}]\\\\\\\Rightarrow \dfrac{x^2+y^2}{x^2+y^2+4x+6y+13}=\dfrac{25}{49}\\\\\Rightarrow 49x^2+49y^2=25x^2+25y^2+100x+150y+325\\\\\Rightarrow 24x^2+24y^2-100x-150y-325=0.$

Thus, the required locus of the point (x, y) is $24x^2+24y^2-100x-150y-325=0.$

Answer 2

Answer:

24x²+24y²-100x-150y-325=0