Find the locus of image of the veriable point `(lambda^(2), 2 lambda)` in the line mirror x-y+1=0, where `lambda` is a peremeter.

If the locus of the image of the point (lambda^(2), 2lambda) in the line mirror x-y+1=0 (where lambda is a parameter) is (x-a)^(2)=b(y-c) where a, b , c in I , then the value of ((a+b)/(c+b)) is equal to

Is there a real value of lambda for which the image of the point (lambda,lambda-1) by the line mirror 3x+y=6 lambda is the point (lambda^(2)+1,lambda) If so find lambda_(.)

The locus of the point x=a+lambda^(2),y=b-lambda where lambda is a parameter is

If the line passing through the points (4,3) and (2,lambda) is perpendicular to the line y=2x+3, then lambda is equal to

Locus of the image of the point (1,1) in the line (5x-2y-7)+lambda(2x-3y+6)=0lambda in R, is

If the points (lambda,-lambda) lies inside the circle x^(2)+y^(2)-4x+2y-8=0, then find the range of lambda.

Locus of the image of the point (1,1) in the line (5x-2y-7)+lambda(2x-3y+6)=0 (lambda in R'), is

If the image of the point M(lambda,lambda^(2)) on the line x+y = lambda^(2) is N(0, 2), then the sum of the square of all the possible values of lambda is equal to

The distance of the point (x,y) from the origin is defined as d = max . {|x|,|y|} . Then the distance of the common point for the family of lines x (1 + lambda ) + lambda y + 2 + lambda = 0 (lambda being parameter ) from the origin is

The equation of the locus of the foot of perpendicular drawn from (5, 6) on the family of lines (x-2)+lambda(y-3)=0 (where lambda in R ) is