# Equation of the locus moving about a line 24

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 $x=-1$. Find the equation of the locus point $P$.

I have tried this question many times before. but i still couldn't get the answer

$$\mathrm{8x^2+9y^2-56x+-18y+89}$$

Really hope u can help me. Thanks

Hint:

we have $$\dfrac{(h-3)^2+(k-1)^2}{(h+1)^2}=9$$

But I think your question actually meant $$\dfrac{(h+1)^2}{(h-3)^2+(k-1)^2}=9$$

The thing you posted as the 'answer' cannot be the answer, because the problem as stated is, 'Find the equation' and the thing you posted is not an equation.

The distance from $A$ to $P$ is $\sqrt{(x-3)^2+(y-1)^2}$. The distance from $A$ to the line $x=-1$ is just $|x+1|$. We want the distance to $A$ to be three times the distance to the line, so: $$\sqrt{(x-3)^2+(y-1)^2} = 3|x+1|$$ Square both sides: $$(x-3)^2+(y-1)^2=9(x+1)^2$$ Expand the squares: $$x^2 - 6x + 9 + y^2 - 2y + 1 = 9x^2 + 18x + 9$$ This is an equation for the locus point, but it's an awkward one. Group like terms and simplify, and you get $$8x^2+24x-y^2+2y-1 = 0$$ where I see only a handful of terms that even appear in the 'answer' that you posted. I suspect that you've mistranscribed both the problem statement and the desired answer.

(Edited to add missing absolute value around (x+1) above)

• You need absolute values in your expression for distance from the point $(x,y)$ to the line $x = -1$: it should be $|x+1|$. (Just consider the point $(-2,0)$.) This also explains why squaring both sides of the equation is permissible, since usually that would introduce spurious solutions. Nov 3 '17 at 0:33
• @Quasicoherent - Right you are. Fixed. Nov 3 '17 at 16:22
• May I ask? Why do we need to actually use the absolute value? Nov 4 '17 at 15:00
• Distance is always non-negative, even if the point is to the left of the line. Nov 5 '17 at 18:58