## Find the equation of the curve

Trigonometric ratios and functions, the unit circle, inverse trig functions, identities, trig graphs, etc.

### Find the equation of the curve

I'm sorry, I know I'm posting a lot, but this is a huge packet, with more than a couple of challenging problems, most likely things I've forgotten along the way going through high school, who knows.

Here's the latest one: A curve is traced by a point P(x,y) which moves such that its distance from point A(-1,1) is three times its distance from point B(2,-1). Determine the equation of the curve.
burnbird16

Posts: 47
Joined: Sat May 22, 2010 1:29 am

burnbird16 wrote: A curve is traced by a point P(x,y) which moves such that its distance from point A(-1,1) is three times its distance from point B(2,-1). Determine the equation of the curve.

You have three points: P = (x, y), A = (-1, 1), and B = (2, -1).

You have the Distance Formula.

You have the relationship d(P,A) = 3*d(P,B), from which you can create an equation.

Square both sides of that equation, simplify, and solve for "y=".

stapel_eliz

Posts: 1717
Joined: Mon Dec 08, 2008 4:22 pm

### Re: Find the equation of the curve

Okay, I'm working through it, but I'm finding y^2's that are kind of difficult to get rid of. I don't know what to do, please help.
burnbird16

Posts: 47
Joined: Sat May 22, 2010 1:29 am

### Re: Find the equation of the curve

Here, maybe this will clarify!

sqrrt((x+1)^2 + (y-1)^2) = 3sqrrt((x-2)^2 + (y+1)^2) <-- Then I squared both sides and got

(x+1)^2 + (y-1)^2 = 9((x-2)^2 + (y+1)^2) Then I simplified.

x^2+2x+1+y^2-2y+1 = 9(x^2-4x+4+y^2+2y+1) Then, I multiplied out the 9 on the right side.

x^2+2x+1+y^2-2y+1 = 9x^2-36x+36+9y^2+18y+9 Then, I moved all the x expressions to the left side, and the y expressions to the right side, to get

-8x^2+38x-35 = 8y^2+20y+8

From there, you can see the problem that arises. I have a full y-expression on one side, and a full x-expression on the other. What do I do?
burnbird16

Posts: 47
Joined: Sat May 22, 2010 1:29 am

I haven't checked your math, but this is one of those instances when the Quadratic Formula can save the day.

Assuming your last line is correct, you then have:

. . . . .$8y^2\, +\, 20y\, +\, (8x^2\, -\, 38x\, +\, 43)\, =\, 0$

That is, you have a quadratic in $x$, with $a\, =\, 8$,$b\, =\, 20$, and $c\, =\, 8x^2\, -\, 38x\, +\, 43$.

It won't be pretty, of course, but the results of the Quadratic Formula will be two equations for "$y=$".

stapel_eliz

Posts: 1717
Joined: Mon Dec 08, 2008 4:22 pm

### Re: Find the equation of the curve

Oh, thank you! But which one will I use? Will I have to check both, see which one works?
burnbird16

Posts: 47
Joined: Sat May 22, 2010 1:29 am

Why can't they both work? (Hint: Draw the picture!)

stapel_eliz

Posts: 1717
Joined: Mon Dec 08, 2008 4:22 pm

### Re:

stapel_eliz wrote:Why can't they both work? (Hint: Draw the picture!)

Oh, wow! You're absolutely right! Thank you!
burnbird16

Posts: 47
Joined: Sat May 22, 2010 1:29 am