a=(a,0) and b=(-a,0) are two fixed points . P is a variable point such that PA and PB intercept a constant length k on a given line y=x find locus of point p

Dear Student,
Let P=(r,s)
eqn of AP: y=s(x-a) /(r-a)
eqn of BP: y=s(x+a)/(r+a)
intersection of y=x and PA give point C by solving tre y=x and PA:
C=(as/(r-a-s),as/(r-a-s))
intersection of y=x and PB give point D by solving tre y=x and PB:
D=(as/(r+a-s),as/(r+a-s))
distance CD= sqrt(2(as/(r-a-s)-as/(r+a-s))^2)
=(sqrt(2)*a^2*s)/(r^2-2sr-a^2+s^2)
given this distance=k
=>k=(sqrt(2)*a^2*s)/(r^2-2sr-a^2+s^2)
=>kr^2+ks^2-sqrt(2)*a^2*s-2krs-ka^2=0
replacing r with x and s with y:
kx^2+ky^2-sqrt(2)*a^2*y-2kxy-ka^2=0 is the required locus of P. [ans]