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A curve is traced by a point $P(x,y)$ which moves such that its distance from the point $A(-1,1)$ is three times its distance from the point $B(2,-1)$. Determine the equation of the curve.

I have only one question. And that is the only thing I need answered at this time. My question to you is, when it says "which moves such that its distance from the point..." by distance, does it mean the slope from $P(x,y)$ to $A(-1,1)$ is three times the distance than from$P(x,y)$ to $B(2,-1)$? Please answer only this and nothing else. I will re-edit with further findings.

Edit: To find the next points would it be logical to use this equation:

$d=distance$ and $P=(x,y)$ $$d(P,(-1,1))=3d(P,(2,-1))$$ $$d\sqrt{(-1\pm x_1)^2+(1\pm y_1)^2}=3d\sqrt{(2\pm x_1)^2+(-1\pm y_1)^2}$$ And from here I would use $P=(x,y)$ and plug in any values of $x$ and $y$ to try and find my equation. Would this be correct?

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You don't need the $d$ any more in the square roots. Remove the $d$, square both sides of your equation, and simplify accordingly. –  J. M. Jul 24 '12 at 6:23
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I would prefer for simplicity $\sqrt{x+1)^2+(y-1)^2}=3\sqrt{(x-2)^2+(y+1)^2}$. Equivalently and much better, square both sides, bring stuff together, there will be a nice surprise. –  André Nicolas Jul 24 '12 at 7:09
    
@AndréNicolas: I think this is right, but I have no idea. I got; $-\dfrac{1}{2}x-4y-\dfrac{11}{4}=0$ –  Austin Broussard Jul 24 '12 at 23:55
    
Sorry, dinner event. Start from my comment above, square both sides we get (x+1)^2+(y-1)^2=9[(x-2)^2+ 9(y+1)^2]$. Expand and simplify. You wil get the equation of a circle. (It will start as $8x^2+8y^2+\cdots=0$. Divide by $8$ to make it look more familiar. –  André Nicolas Jul 25 '12 at 3:17
    
@AndréNicolas sorry I was on a little vacation. Came back to this problem and I got the equation: $0=8(x^2-y^2-\frac{1}{4}x+2y+5\frac{1}{4})$. I do not believe this is correct in any way. –  Austin Broussard Aug 4 '12 at 5:41

2 Answers 2

At all points on the curve the distance from $P$ to $A$ should be three times the distance from $P$ to $B$.

I recommend first sketching this by hand. Then use the distance formula to find the equation of the curve. Finally, make sure both answers agree!

Let $d(P,Q)$ represent the distance between points $P$ and $Q$. The problem statement tells us that $$d(P,A) = 3 d(P,B).$$ Plug in $(x,y)$ for $P$ and the given values of $A$ and $B$ to find the equation for the curve.

Hint: After some algebra you will find it is one of the conic sections.

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@AustinBroussard: Glad to help. Slope is not distance. Use the distance formula. –  user26872 Jul 24 '12 at 4:53
    
So this means that I have to plot $P$ in such a way that anywhere on the curve, the distance from $P$ to $A$ is exactly three times the distance than from $P$ to $B$? –  Austin Broussard Jul 24 '12 at 4:55
    
@AustinBroussard: You've got it. :-) –  user26872 Jul 24 '12 at 4:56
    
So this means that the above statement entails the point $P$ has to be closer to point $B$? Meaning in the fourth quadrant? –  Austin Broussard Jul 24 '12 at 4:59
    
@AustinBroussard: Mostly ... –  user26872 Jul 24 '12 at 5:05

The distance between two points is not the slope. For example, the distance between $(1,2)$ and $(6,9)$ is $\sqrt{74}$. No more, since you asked for minimal help. And congratulations for that!

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Congratulations on not asking for much help? –  Austin Broussard Jul 24 '12 at 4:59
    
Yes, too many people want full solutions, and lose an opportunity to learn. –  André Nicolas Jul 24 '12 at 5:07
    
Right, I really hate that! But I have another question: supposed the distance from $P\rightarrow A=15$, would this make the distance from $P\rightarrow B=5$? –  Austin Broussard Jul 24 '12 at 5:08
    
Yes, it would. ${}{}{}{}{}{}$ –  André Nicolas Jul 24 '12 at 5:10
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@AustinBroussard: Yes, that is good. –  André Nicolas Jul 24 '12 at 5:31

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