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.

  • $\begingroup$ 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

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 .

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.