Find the relation between x and y such that the distance of point p(x,y) and (3,6)is twice the distance between the points p(x,y) and (-3,4)
Answers
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)
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)
hope it helps