Question

A curve is traced by a point P(x,y) which moves such that it's distance from the point A(1,-2) is twice its distance from the point B(4,-3). Determine the equation of the curve.

Answer 1

Do you know the distance formula?

Answer 2

Yes, but I don't know how to use the information within that formula to find what it is looking for

Answer 3

*asking for

Answer 4

Well, a point P on the curve is (x,y). The distance formula says that \[\overline{AP}=\sqrt{(x-1)^2+(y+2)^2}\]and\[\overline{BP}=\sqrt{(x-4)^2+(y+3)^2}\]Does that make sense?

Answer 5

Ok I understand, where do I go from there?

Answer 6

\[\overline{AP}=2\overline{BP}\]We can already tell that this equation involves nasty square roots. How do you get rid of square roots? Do you know?

Answer 7

Square them? Or conjugate them

Answer 8

Conjugating them is a little more complicated, but it does the exact same thing. So we'll just square both sides.\[\overline{AP}^2=\left ( 2\overline{BP}\right ) ^2=4\overline{BP}^2\]\[(x-1)^2+(y+2)^2=4\left ( (x-4)^2+(y+3)^2\right ) =4(x-4)^2+4(y+3)^2\]\[\left ( x^2-2x+1\right ) +\left ( y^2+4x+4\right ) =\left ( 4x^2-32x+64\right ) +\left ( 4y^2+24y+36\right ) \]Do you understand it?

Answer 9

I'm getting there, now can you explain why it all equals 4bp squared

Answer 10

Well, AP is two times BP, so AP squared is four times BP squared.\[\overline{AP}=2\overline{BP}\]If you square both sides you get\[\left ( \overline{AP}\right ) ^2=\left ( 2\overline{BP}\right ) ^2\]\[\left ( \overline{AP}\right ) ^2=(2)^2\left ( \overline{BP}\right ) ^2\]\[\left ( \overline{AP}\right ) ^2=4\left ( \overline{BP}\right ) ^2\]

Answer 11

Ok I got it, now you have it split into the four quadratic equations, what happens next?

Answer 12

Well, the rest of the problem is pretty hard. First, you need to collect like terms. Lets call "Collecting like terms" CLT for short. CLT has multiple steps.

CLT1: collect all of the constant terms on one side of the equation.\[x^2-2x+\underset{\uparrow}{1}+y^2+4y+\underset{\uparrow}{4}=4x^2-32x+\underset{\uparrow}{64}+4y^2+24y+\underset{\uparrow}{36}\]\[x^2-2x+y^2+4y+\underset{\uparrow}{5}=4x^2-32x+4y^2+24y+\underset{\uparrow}{100}\]\[x^2-2x+y^2+4y=4x^2-32x+4y^2+24y+\underset{\uparrow}95\]Does that seem like a good first step to you?

Answer 13

Yea, I got it, you add all the numbers without variables together

Answer 14

So, actually we just want all the terms on one side of the equation (you'll see why later). So for CLT2, we put all the variables to the first power on the right hand side (since we already have all our constants on the right hand side)\[x^2-\underset{\uparrow}{2x}+y^2+\underset{\uparrow}{4y}=4x^2-\underset{\uparrow}{32x}+4y^2+\underset{\uparrow}{24y}+95\]\[x^2+y^2=4x^2-32x+\underset{\uparrow}{2x}+4y^2+24y-\underset{\uparrow}{4y}+95\]\[x^2+y^2=4x^2-30x+4y^2+20y+95\]And then for CLT3, we gather together the squared terms on the RHS, and we get this:\[0=3x^2-30x+3y^2+20y+95\]So then I'm guessing you want to solve for y, so I'll rearrange it in the following way:\[0=3y^2+20y+\left ( 3x^2-30x+95\right ) \]Which is the same thing as \(0=ay^2+by+c\) with \(a=3\), \(b=20\), \(c=3x^2-30x+95\).

Answer 15

So, after the quadratic formula (since we have \(0=ay^2+by+c\) ), and some simplification, we get:\[y=-\frac{10}{3}\pm 2\sqrt{\frac{40}{9}-(x-5)^2}\]Which is actually an ellipse (oval) of some sort. Take from the solution what you will.

Answer 16

Thank you so much. I appreciate your step by step walk through