${y^2} = y\left( {-\dfrac{1}{{{y_1}}}} \right) \cdot \left[ {y\left( {-\dfrac{1}{{{y_1}}}} \right) + 2x} \right] \Rightarrow {y^2}y_1^2 = {y^2} - 2xy{y_1}$

${y^2} = {y^2}y_1^2 + 2xy{y_1}$

$y^2 = y y_1 [yy_1 + 2x]$

Which is same as in equation (3)
So, differential equation of given curve and differential equation of its orthogonal trajectaries are same. So, the family of curves is self orthogonal.

Example 3 :

Find the orthogonal trajectories of the family of curves $\dfrac{{{x^2}}}{{{a^2} + \lambda }} + \dfrac{{{y^2}}}{{{b^2} + \lambda }} = 1$ , where $\lambda$ is a parameter.

Solution :

We are given $\dfrac{{{x^2}}}{{{a^2} + \lambda }} + \dfrac{{{y^2}}}{{{b^2} + \lambda }} = 1$

Differentiate both sides w.r.t x

$\dfrac{{2x}}{{{a^2} + \lambda }} + \dfrac{{2y}}{{{b^2} + \lambda }} \cdot \dfrac{{dy}}{{dx}} = 0 \Rightarrow \dfrac{x}{{{a^2} + \lambda }} + \dfrac{y}{{{b^2} + \lambda }} \cdot \dfrac{{dy}}{{dx}} = 0$ …(1)

On solving $x({b^2} + \lambda ) + y({a^2} + \lambda ) \dfrac{{dy}}{{dx}} = 0$

$\lambda \left[ {x + y \dfrac{{dy}}{{dx}}} \right] = -\left[ {{b^2}x + {a^2}y \dfrac{{dy}}{{dx}}} \right]$

$\lambda = - \dfrac{{\left[ {{b^2}x + {a^2}y \dfrac{{dy}}{{dx}}} \right]}}{{x + y \dfrac{{dy}}{{dx}}}}$

$a^2 + \lambda = a^2 - \dfrac{b^2 x + a^2 y \ \ dy/dx}{x + y \ dy/dx} = \dfrac{(a^2 - b^2)x}{x + y \ dy/dx}$

$b^2 + \lambda = b^6 - \left [ \dfrac{b^2 x + a^2 y \dfrac{dy}{dx}}{x + y \dfrac{dy}{dx}} \right ] = \dfrac{(-a^2 + b^2)y (dy/dx)}{x + y \ dy/dx}$

Now on substituting values of $a^2 + \lambda$ and $a^2 + \lambda$ in (i)

$x^2 \cdot \dfrac{x + y dy/dx}{(a^2 - b^2)x} + y^2 \cdot \dfrac{x + y dy/dx}{(-a^2 + b^2)y \dfrac{dy}{dx}} = 0$

$\left [x + y \dfrac{dy}{dx} \right ] \left [ \dfrac{x}{a^2 - b^2} - \dfrac{y}{(a^2 - y^2) \dfrac{dy}{dx}} \right ] = 1$

$\left [x + y \dfrac{dy}{dx} \right ] \left [x - y \dfrac{dx}{dy} \right ] = (a^2 - b^2),$ which is the differential equation is given family. …(2)

Now replace $\dfrac{dy}{dx}$ with $-\dfrac{dx}{dy}$ to obtain differential equation of orthogonal trajectory

$\left [x - y \dfrac{dx}{dy} \right ] \left [x + y \dfrac{dy}{dx} \right ] = (a^2 - b^2 )$

Differential equation (2) and (3) are same, which gives the differential equation of family. It’s orthogonal trajectories are same. So the family of curves are self orthogonal.

Example 4 :

Show that the families of curves given by the equation $r = a(1 - \cos \theta)$ and $r = b(1 + \cos \theta)$ intersect orthogonally.

Solution :

Here we have to show that the family of orthogonal trajectory of the family of curves $r = a(1 - \cos \theta)$

$r = b(1 + \cos \theta )$

We have $r = a(1 - \cos \theta )$ …(1)

On taking logrithm both sides

$\log r = \log a + \log(1 - \cos \theta)$

Differentiate both sides w.r.t $\theta$

$\dfrac{1}{r} \cdot \dfrac{dr}{d \theta} = \dfrac{\sin \theta}{1 - \cos \theta}$ …(2)

which is free from parameter. So, it is the differential equation of given family.

Now replace $\dfrac{dr}{d \theta}$ with $- r^2 \dfrac{d \theta}{d r}$ in (2)

$\dfrac{1}{r}\left( {-{r^2}\dfrac{{d\theta }}{{dr}}} \right) = \dfrac{{\sin \theta }}{{1 - \cos \theta }} = 2\dfrac{{\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}}}{{2{{\sin }^2} \dfrac{\theta }{2}}} = \cot \dfrac{\theta }{2}$

$- r \dfrac{d \theta}{dr} = \cot \dfrac{\theta}{2} \Rightarrow \dfrac{dr}{r} = -\tan \dfrac{\theta}{2} d \theta$

On integrating $\log r = 2 \log \cos \dfrac{\theta}{2} + \log c$

$\log r = \log \left( {c \ {{\cos }^2} \dfrac{\theta }{2}} \right)$

$r = c\left( {\dfrac{{1 + \cos \theta }}{2}} \right)$

$r = b(1 + \cos \theta )$

which is the required orthogonal trajectory.

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