Question

find the orthogonal trajectories of the family of curve where cy square =xcube are arbitrary constant?​

Answer 1

Answer:

☆Solution☆ :-

Given quation of the family of curves is 

$\bf \: c{y}^{2} = {x}^{3}$

can be rewritten as

$\bf \: {y}^{2} = \dfrac{1}{c} {x}^{3} .........(1)$

c is an arbitrary constant.

Now, differentiating the equation with respect to x gives,

$\bf\implies \:2cy\dfrac{dy}{dx} = {3x}^{2}$

$\bf\implies \: \dfrac{2y}{3 {x}^{2} } \dfrac{dy}{dx} = \dfrac{1}{c}$

Put this value in equation (1), we get

$\bf\implies \: {y}^{2} = \dfrac{2y}{3 {x}^{2} } \dfrac{dy}{dx} \times {x}^{3}$

$\bf\implies \:3y = 2x\dfrac{dy}{dx} ......(2)$

$\bf \:Now, \: to \: find \: the \: orthogonal \: trajectories$

$\bf \:we \: need \: to \: replace  \dfrac{dy}{dx} \: by \: \dfrac{ - dx}{dy}$

So, equation (2) can be rewritten as

$\bf\implies \:3y = 2x\dfrac{ - dx}{dy}$

$\bf\implies \:3ydy = - 2xdx$

Integrating both sides, we get

$\bf\implies \:∫3ydy = - ∫2xdx$

$\bf\implies \: \dfrac{ {3y}^{2} }{2} = - {x}^{2} + c$

which is equation of the orthogonal trajectories of the given family of curves.

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