Question Based on orthogonal Trajectories

Exact Differential Equation and Orthogonal Trajectories

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

Find the orthogonal trajectories of xy=c

Find the orthogonal trajectories of xy=c

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

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

Illustration based upon Orthogonality Pole-Polar and Parametric Equation OF Circle

Illustration based upon Orthogonality Pole-Polar || Parametric Equation OF Circle