1. ## locus- precalculua

can someone help me plz. find the locus of a point P which moves so that its distance from A(8,0) is twice its distance from the line x=2

2. Let $\displaystyle P(x,y)$.
The distance from P to A is $\displaystyle \sqrt{(x-8)^2+y^2}$.
The distance from P to the line is $\displaystyle |x-2|$.
Now, we have
$\displaystyle \sqrt{(x-8)^2+y^2}=2|x-2|$.
Square both sides:
$\displaystyle (x-8)^2+y^2=4(x-2)^2\Rightarrow 3x^2-y^2=48$
Divide both sides by 48:
$\displaystyle \displaystyle \frac{x^2}{16}-\frac{y^2}{48}=1$.
So, the locus is a hyperbola.

3. hey thanks so much hun

4. ## hey

we also have to make a sketch. can u plz help me with that?

5. Here's the sketch.

6. ## hey

i don't understand how u got the answer for the question. can u break down the steps plz