Find the orthogonal trajectories of `xy=cdot`

Find the orthogonal trajectories of family of curves x^(2)+y^(2)=cx

Find the orthogonal trajectory of y^(2)=4ax (a being the parameter).

The equation to the orthogonal trajectories of the system of parabolas y=ax^(2) is

Any curve which cuts every member of a given family of curve is called an orthogonal trajectory of family. To get this, we replace (dy)/(dx)by -(dx)/(dy) in differential equation. The orthogonal trajectories for y=ce^(-x) represent

Any curve which cuts every member of a given family of curve is called an orthogonal trajectory of family. To get this, we replace (dy)/(dx) by -(dx)/(dy) in differential equation. The orthogonal trajectories for y=cx^(2) , where c is a constant is

STATEMENT-1 : The orthogonal trajectory of a family of circles touching x-axis at origin and whose centre the on y-axis is self orthogonal. and STATEMENT-2 : In order to find the orthogonal trajectory of a family of curves we put -(dx)/(dy) in place of (dy)/(dx) in the differential equation of the given family of curves.

A curve 'C' with negative slope through the point(0,1) lies in the I Quadrant. The tangent at any point 'P' on it meets the x-axis at 'Q'. Such that PQ=1 . Then The orthogonal trajectories of 'C' are

If the orthogonal trajectories of the family of curves e^(n-1)x=y^(n) .(where e is a parameter) always passes through (1, e) and (0, 0) then value of n is