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Topic: Tough locus
Replies: 10   Last Post: Jun 1, 2013 5:39 PM

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Avni Pllana

Posts: 546
Registered: 12/6/04
Re: Tough locus
Posted: May 23, 2013 3:26 AM
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>>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.

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 x-axis. Similarly we have

tan(A)=y/(x+c), (2)

tan(B)=y/(x-c). (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(A-P)=(y-(x+c)*y')/(x+c+y*y'), (5)

tan(bet)=tan(P-B)=-(v-(x-c)*y')/(x-c+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

[(x-c)^2+y^2]*[x+c+y*y']^2=m^2*[(x+c)^2+y^2]*[x-c+y*y']^2 . (8)

For the special case m = 1, from (8) we have

y'*(x^2-y^2-c^2)+x*y*(y'^2-1) = 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^2-x^2), and c=sqrt(a^2-b^2) .

Best regards,

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