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A point $P(x,y)$ moves in such a way that its distance from the point $A(3,1)$ is always three times its distance from the straight line $x=-1$.

(a) Find the equation of the locus of the moving point P.

(b) Determine whether the locus of the point P will intersect the straight line $y=-1$.

I have solved (a) but was unable to solve (b). I have tried to substitute $y=-1$ into the locus equation.

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  • $\begingroup$ on the line $y=-1$, $x$ is zero so put $x=0,y=-1$ in the equation of locus, if this point satisfies the equation then Locus will intersect the line $y=-1$ $\endgroup$ Aug 16 '15 at 8:43
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a) The distance of the point $P(x, y)$ from the point $A(3, 1)$ is such that $$\sqrt{(x-3)^2+(y-1)^2}=3\times \frac{|x+1|}{\sqrt{1^2+(0)^2}}$$ $$(x-3)^2+(y-1)^2=9(x+1)^2$$ $$x^2-6x+9+y^2-2y+1=9(x^2+2x+1)$$ $$8x^2-y^2+24x+2y-1=0$$ $$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{equation of the locus of point P:}\ 8x^2-y^2+24x+2y-1=0}}$$

Above is the locus of the point $P(x, y)$

b) setting $y=-1$ in the above equation, we get $$8x^2-(-1)^2+24x+2(-1)-1=0$$ $$2x^2+6x-1=0$$ Checking the nature of roots of above quadratic equation by using discriminant $\Delta$, as follows $$\Delta=B^2-AC=(6)^2-4(2)(-1)=44>0$$ Above positive value of the discriminant shows that there are two distinct real roots i.e. the locus of the point P intersects the straight line: $y=-1$ at two different points .

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  • $\begingroup$ Can this be done using b2-4ac? $\endgroup$
    – Gunners98
    Aug 16 '15 at 8:56
  • $\begingroup$ No need to calculate the discrimant: it is positive since the additive constant is negative. $\endgroup$ Aug 16 '15 at 9:13
  • $\begingroup$ Yes, you are absolutely right $\endgroup$ Aug 16 '15 at 9:15

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