Home
Class 12
MATHS
Find the locus of image of the veriable ...

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.

Text Solution

AI Generated Solution

To find the locus of the image of the variable point \((\lambda^2, 2\lambda)\) in the line mirror \(x - y + 1 = 0\), we will follow these steps: ### Step 1: Identify the point and the line Let the variable point \(P\) be given by the coordinates \(P(\lambda^2, 2\lambda)\). The line mirror is given by the equation \(x - y + 1 = 0\). ### Step 2: Find the coefficients of the line The line can be rewritten in the standard form \(Ax + By + C = 0\), where: - \(A = 1\) ...
Promotional Banner

Topper's Solved these Questions

  • STRAIGHT LINES

    CENGAGE ENGLISH|Exercise EXAMPLE|12 Videos
  • STRAIGHT LINES

    CENGAGE ENGLISH|Exercise CONCEPT APPLICATION EXERCISE 2.1|23 Videos
  • STRAIGHT LINE

    CENGAGE ENGLISH|Exercise Multiple Correct Answers Type|8 Videos
  • THEORY OF EQUATIONS

    CENGAGE ENGLISH|Exercise JEE ADVANCED (Numerical Value Type )|1 Videos

Similar Questions

Explore conceptually related problems

If the points A(lambda, 2lambda), B(3lambda,3lambda) and C(3,1) are collinear, then lambda=

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

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

Find the centre and the radius of the circles 2x^(2) + lambda xy + 2y^(2) + (lambda - 4) x+6y - 5 = 0 , for some lambda

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

If the line y=2x+lambda be a tangent to the hyperbola 36x^(2)-25y^(2)=3600 , then lambda is equal to

If the tangents from the point (lambda, 3) to the ellipse x^2/9+y^2/4=1 are at right angles then lambda is

Find the value of lambda so that the line 3x-4y= lambda may touch the circle x^2+y^2-4x-8y-5=0

The locus of the image of the point (2,3) in the line (x-2y+3)+lambda(2x-3y+4)=0 is (lambda in R) (a) x^2+y^2-3x-4y-4=0 (b) 2x^2+3y^2+2x+4y-7=0 (c) x^2+y^2-2x-4y+4=0 (d) none of these