The point p(x,y) moves so that its distance from the point k(2,5) is twice its distance from the line x=-1. Find the equation of the locus of P.

Question

This question is from the Yr 11 Cambridge 3U Enhanced textbook and there are answers in the book but I don't know how to do it.  

Answer 1

The distance from (x,y) to (2,5) is given by distance formula:

d = √[(x-2)²+(y-5)²]

 

The distance from (x,y) to the line x=-1 will taken from

the perpendicular distance as shortest which will be a

horizontal distance since x=-1 is a vertical line.

This distance will be the difference in the x coordinates

stated as a positive or... |x-(-1)| or |x+1|

 

Since we want the distance from (x,y) to (2,5) to be

twice the distance from (x,y) to x=-1 we have...

 

√[(x-2)²+(y-5)²] = 2|x+1|

 

square both sides... Note that since squaring (x+1) will

automatically result in a positive we will no longer need the

absolute value sign

 

(x-2)²+(y-5)² = 4(x+1)²

 

x²-4x+4+y²-10y+25 = 4(x²+2x+1)

x²-4x +y²-10y+29 = 4x²+8x+4

 

Get everything to one side of the = sign except the constant.

I will subtract the left from the right just because

I prefer my x² to be a positive coefficient....

 

3x² + 12x - y² + 10y = 25

 

Note this is becoming the equation of a hyperbola.

We want standard form (x-h)²/a² - (y-k)²/b² = 1 :

We need to do some complete the square work here.

 

3(x²+4x) - (y²-10y) = 25

3(x²+4x+4) - (y²-10y + 25) = 25 + 12 - 25

 

3(x+2)² - (y-5)² = 12

 

(x+2)²/4 - (y-5)²/12 = 1

 

We have a hyperbola centered at (h,k) or (-2,5)