
Re: Tough locus
Posted:
May 23, 2013 3:26 AM


>>Find locus of a point P moving along a curve whose tangent >>is inclined to two rays from A and B ( distance 2 c apart) >>at angles satisfying cos(alpha)/cos{beta) = m ( a >>constant). Special case is ellipse for m = 1. >> >>TIA >>Narasimham
Hi Narasimham,
welcome back to the good old geometry puzzles forum. I tried to solve the problem as follows:
Let A=[c,0], B=[c,0], P=[x,y]. We have
tan(P)=dy/dx=y', (1)
where 'P' denotes the angle between the tangent at point P and the positive xaxis. Similarly we have
tan(A)=y/(x+c), (2)
tan(B)=y/(xc). (3)
For an arbitrary angle 'phi' we have
cos(phi)=1/sqrt(1+(tan(phi))^2). (4)
Let 'alf', 'bet' be the angles between the tangent at point P and the rays AP, BP respectively. From (1),(2),(3) we have
tan(alf)=tan(AP)=(y(x+c)*y')/(x+c+y*y'), (5)
tan(bet)=tan(PB)=(v(xc)*y')/(xc+y*y'). (6)
From the condition of the problem we have
cos(alf)/cos(bet) = m . (7)
Now from (4),(5),(6),(7) we obtain the following differential equation
[(xc)^2+y^2]*[x+c+y*y']^2=m^2*[(x+c)^2+y^2]*[xc+y*y']^2 . (8)
For the special case m = 1, from (8) we have
y'*(x^2y^2c^2)+x*y*(y'^21) = 0 . (9)
MATLAB was not able to find an explicit solution for (9).
However, (9) is satisfied if we substitute the equation of the ellipse
y = b/a*sqrt(a^2x^2), and c=sqrt(a^2b^2) .
Best regards, Avni

