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Let the trajectories cut the crve of giv...

Let the trajectories cut the crve of given family at an angle `alpha` where `tna alpha = k`.
The slope `(dy)/(dx) = tan Psi` (of the tangent to a member of the family and the slope `(dy_(T))/(dx) = tan Phi` to the isogonal trajectory are connected by the relationship

`tan phi = tan (Psi - alpha) = (tan Psi - tan alpha)/(1+ tan alpha tan Psi)`
i.e., `(dy)/(dx) = (((dy_(T))/(dx))-k)/(k(dy_(T))/(dx)+1)`
Substituting this expression into equation, (l') and dropping the subscript T, we obtain the differential equation of isogonal trajectories.
The isogonal trajectories of a family of straight lines y = c, that cuts the given family at angle `alpha`, the tangent of which is k, is

A

(a)y = kx

B

(b)`y = k tan alpha x`

C

(c)`y = k cot 2 x`

D

(d)y = cx

Text Solution

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The correct Answer is:
A
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