Question

find orthogonal trajectories of the parabola family ay²=x³​

Answer 1

Working Rule :-

1. Differentiate the given curve f(x, y, a)

2. Obtain the value of arbitrary constant a.

3. Substituting the value of arbitrary constant 'a' in f(x, y, a) to eliminate a.

4. Then replace dy/dx by - dx/dy.

5. If we get the same equation back, curve is called Self - Orthogonal otherwise integrating both sides.

Let's solve the problem now!!!!

$\green{\large\underline{\bf{Solution-}}}$

The given equation is

$\bf :\longmapsto\: {ay}^{2} = {x}^{3} - - - (1)$

Differentiating both sides w. r. t. x, we get

$\rm :\longmapsto\:\dfrac{d}{dx} {ay}^{2} = \dfrac{d}{dx} {x}^{3}$

$\rm :\longmapsto\: \: 2ay\dfrac{dy}{dx} = 3 {x}^{2}$

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

On substituting the value of 'a' in equation (1), we get

$\rm :\longmapsto\:\dfrac{3 {x}^{2} }{2y\dfrac{dy}{dx}} {y}^{2} = {x}^{3}$

$\rm :\longmapsto\: {3x}^{2}y\dfrac{dx}{dy} = 2{x}^{3}$

$\rm :\longmapsto\: 3y\dfrac{dx}{dy} = 2x$

$\red{\bf :\longmapsto\:On \: replacing \: \dfrac{dx}{dy} \: by \: - \dfrac{dy}{dx}, \: we \: get}$

$\rm :\longmapsto\: - \: 3y\dfrac{dy}{dx} = 2x$

$\rm :\longmapsto\: - 3ydy = 2xdx$

On integrating both sides, we get

$\displaystyle \: \rm :\longmapsto\: - \int 3ydy = \int 2xdx$

$\rm :\longmapsto\: - \dfrac{3 {y}^{2} }{2} = {x}^{2} + c$

$\rm :\longmapsto\: - 3 {y}^{2} = 2{x}^{2} + 2c$

$\rm :\longmapsto\: 3 {y}^{2} + 2{x}^{2} + 2c = 0$

$\rm :\longmapsto\: 3 {y}^{2} + 2{x}^{2} +d = 0 \: \: \: \: where \: d \: = \: 2c$

Thus,

$\: \: \: \underbrace{\boxed{ \bf \: {2x}^{2} + {3y}^{2} + d = 0 \: is \: required \: equation}}$

Basic Formula's Used :-

$\boxed{ \sf \: \dfrac{d}{dx} {x}^{2} = 2x}$

$\boxed{ \sf \: \dfrac{d}{dx} {x}^{n} = {nx}^{n - 1}}$

$\boxed{ \sf \: \int \: {x}^{n}dx = \dfrac{ {x}^{n + 1} }{n + 1} + c}$

Answer 2

Answer: The equation 2x²+3y²+d is the orthogonal trajectories of the parabola family.

Orthogonal Trajectories: A curve that crosses any curve of a given pencil of (planar) curves orthogonally is called an orthogonal trajectory in mathematics.

Step-by-step explanation:

Step 1: Given data

The equation of the parabola $ay^2= x^3---- > 1$

where a is a constant.

Step 2: Removing the constant a from equation 1

Differentiating 1 w.r.t. x

$\frac{d}{dx} ay^2=\frac{d}{dx}x^2\\2ay\frac{dy}{dx} = 3x^2\\a = \frac{3x^2}{2y\frac{dy}{dx}}----- > 2$

On Substituting the value of 'a' in equation 1

$\frac{3x^2}{2y\frac{dy}{dx}}y^2=x^3\\3y\frac{dx}{dy}=2x$

For orthogonal trajectories $\frac{dx}{dy} = -\frac{dy}{dx}$

$-3y\frac{dy}{dx}=2x\\-3ydy=2xdx---- > 3$

Step 3: Integrating equation 3 to get orthogonal trajectories

$\int {-3y} \, dy = \int {2x} \, dx\\\frac{-3}{2}y^2 = x^2 + c\\Therefore\hspace{2}, 3y^2+2x^2+d = 0 \hspace{10} where \hspace{2},d = 2c$

The above equation is the equation for all possible orthogonal trajectories for family of the given parabola