Vikas TU
Last Activity: 4 Years ago
P lies on y=x , hence its coordinates are P(a,a)
Q lies on y=2x , hence its coordinates are Q(b,2b)
Given that |PQ|=4
Hence (a−b)^2+(a−2b)^2=42=16 ………………(1)
Now let M be the mid point of PQ. M has coordinates
a+b^2,a+2b^2
To trace M’s locus, we assign x and y coordinates as :
x=a+b/2 ………………(2)
and
y=a+2b/2 ………………(3)
Solving for a and b from eqn(2) and eqn(3) above we get,
b=2y−2x ………………(4)
a=4x−2y ………………(5)
Substituting eqn(4) and eqn(5) into eqn(1) gives the desired locus. That’s :
16=(6x−4y)2+(8x−6y)^2
Hence
4=(3x−2y)^2+(4x−3y)^2
Hence the equation of the locus is :
25x^2−36xy+13y^2−4=0
