finding the equation of a locus: P is always half as far....

[Joyce]

Question: Point P moves so it is always one-half as far from point A (-5, 0) as it is from point B (5, 0)

So I used the distance formulas and divided BP by 0.5. It came to:

. . .x^2 + 10x + 25 + y^2 = 2x^2 - 20x + 50 + 2y^2

The answer at the back of the book is:

. . .3x^2 + 3y^2 + 50x + 75 = 0

How do I get that?

 

[pka]

You are making exactly the same mistake as you made before:
2AP=PB so 4AP<SUP>2</SUP>=PB<SUP>2</SUP>.

 

[soroban]

Re: finding the equation of a locus

Hello, Joyce!

You simply must read the problems carefully . . .

Point P moves so it is always one-half as far from point A (-5, 0) as it is from point B (5, 0)

So I used the distance formulas and divided BP by 0.5. . . .divided by 0.5 . . . why?

It is unfortunate that the English language allows a variety of phrases to have the same meaning
\(\displaystyle \;\;\)and they can be readily misread.

The first sentence says: "\(\displaystyle PA\) is one-half of \(\displaystyle PB\)": \(\displaystyle \:\overline{PA}\;=\;\frac{1}{2}\overline{PB}\)

So we have: \(\displaystyle \,\sqrt{(x\,+\,5)^2\,+\,y^2}\;=\;\frac{1}{2}\sqrt{(x\,-\,5)^2\,+\,y^2}\)

And remember: the Distance Formula has a square root in it.