The slope of the tangent is dy/dx and the slope of line joining the

1. i) Find curves for which tangent at a point always perpendicular to the line joining the point to the origin The slope of the tangent is dy/dx and the slope of line joining the point (x,y) to the origin is y/x and since these lines are given to be orthogonal dy/dx=-x/y Integrating x2y2=a which respresents a family of concentric circle.

2. (ii) Find curves for which the projection of the normal on the x-axis is of constant length. This condition gives y dy/dx=k Integrating y2=2kxA which represents a family of parabolas, all with the same axis and same length of latus rectum.

3. (iii) Find curves for which tangent makes a constant angle with the radius vector. Here it is convenient to use polar coordinates and t he conditions of the problem gives r d/dr=tan Integrating r=Ae cot  which represents a family of equiangular spirals.

4. 2.2 Orthogonal Trajections Let f(x,y,a)=0----------------(1) represent a family of curves,one curve for each value of the parameters a. Differentiating, we get f/ x+ f/ y dy/dx=0--------(2)

5. Eliminating a between (1) and (2),we get a differential equation of the first order (x,y,dy/dx)=0 of which (1) is the general solution. Now we want a family of curves cutting every member of (1) at right angle at all points of intersection. At a point of intersection of the two curves,x,y are the same but the slope of the second curve is negative reciprocal of the slope of the first curve.As such differential equation

6. Of the family of orthogonal trajectories is (x,y,-1/ dy/dx)=0 Integrating, we get ---- g(x,y,b)=0 which give the orthogonal trajectories of the family (1) Let the original family be y=mx, when m is a parameter then dy/dx=m and eliminating m, we get the differential equation of this concurrent family of straight lines as y/x=dy/dx To get the orthogonal trajectories, we replace dy/dx by-1/ dy/dx to get y/x=-1/ dy/dx

7. Integrating x2+y2=a2 which gives the orthogonal trajectories as concentric circle. Find the orthogonal trajectories of the family of confocal conics x2/a2++y2/b2+=1 where  is a parameter.Differentiating,we get x2/a2++y2/b2+ dy/dx=0 Eliminating  ,we get (x-y)(x+ y)= (a2+b2) ; =dy/dx To get the orthogonal trajectories, we replace by -1/ to get (-x/;-y)(x- y/)=-1/ (a2-b2 ) or (x-y)(x+ y)= (a2+b2)

8. As such the family of confocal conics is self-orthogonal, i.e. for every conic of the family,there is another with same focii which cuts it at right angles. One family consists of confocal ellipses and the other consists of confocal hyperbolas with the same focii. In polar coordinates after getting the differential equation of the family of curves,we have to replace r d/dr by -1/ r d/dr and then integrate the resulting differential equation.

9. Then if the original family is r=2a cos with a0 as a parameter, its differential equation is obtained by eliminating a and dr/d=-2asin  to get r dr/d=-cot  Replacing r dr/d by -( r dr/d)-1,we get r dr/d=tan  Integrating we get r=2bsin 

10. The orthogonal trajectories are shown The circles of both families pass through the origin, but while the centre of one family lie on x-axis, the centres of the orthogonal family lie on y-axis.

11. The End~