Answer to Question #250950 in Analytic Geometry for Heart

Question

Find the equation of the locus of a point so that the square of its distance from (3, -3) is always numerically equal to the slope of the line joining it to the same point.

Answer 1

Given,

Co-ordinate of the point is (3,-3)

Let the co-ordinate of the general points be (x, y)

So, distance between the points are$(d)=\sqrt{(x-3)^2+(y+3)^2}$

Slope of the line passing from this (3,-3) and (x,y)

$\tan \theta=\frac{y+3}{x-3}$

Now, as per the condition given in the question,

$\Rightarrow d^2=\tan\theta$

Now, substituting the values,

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

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

$\Rightarrow (x^2+y^2-6x+6y+18)(x-3)=y+3$

$\Rightarrow x^3-6x^2+xy^2+6xy+18x-3x^2-3y^2+18x-18y-72=y+3$

$\Rightarrow x^3-9x^2+36x+75+y^2(x-3)+y(19+6x)=0$