Question #250950

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.


1
Expert's answer
2021-10-14T11:59:04-0400

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)=(x3)2+(y+3)2(d)=\sqrt{(x-3)^2+(y+3)^2}

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

tanθ=y+3x3\tan \theta=\frac{y+3}{x-3}

Now, as per the condition given in the question,

d2=tanθ\Rightarrow d^2=\tan\theta

Now, substituting the values,


(x3)2+(y+3)2=y+3x3\Rightarrow (x-3)^2+(y+3)^2=\frac{y+3}{x-3}


(x2+96x+y2+9+6y)(x3)=y+3\Rightarrow (x^2+9-6x+y^2+9+6y)(x-3)=y+3


(x2+y26x+6y+18)(x3)=y+3\Rightarrow (x^2+y^2-6x+6y+18)(x-3)=y+3


x36x2+xy2+6xy+18x3x23y2+18x18y72=y+3\Rightarrow x^3-6x^2+xy^2+6xy+18x-3x^2-3y^2+18x-18y-72=y+3


x39x2+36x+75+y2(x3)+y(19+6x)=0\Rightarrow x^3-9x^2+36x+75+y^2(x-3)+y(19+6x)=0


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