Orthogonal trajectories of family of the curve `x^(2/3)+y^2/3=a^((2/3))` , where `a` is any arbitrary constant, is (a) `( b ) (c) (d) x^(( e ) (f) (g)2/( h )3( i ) (j) (k))( l )-( m ) y^(( n ) (o) (p)2/( q )3( r ) (s) (t))( u )=c (v)` (w) (b) `( x ) (y) (z) x^(( a a ) (bb) (cc)4/( d d )3( e e ) (ff) (gg))( h h )-( i i ) y^(( j j ) (kk) (ll)4/( m m )3( n n ) (oo) (pp))( q q )=c (rr)` (ss) (c) `( d ) (e) (f) x^(( g ) (h) (i)4/( j )3( k ) (l) (m))( n )+( o ) y^(( p ) (q) (r)4/( s )3( t ) (u) (v))( w )=c (x)` (y) (d) `( z ) (aa) (bb) x^(( c c ) (dd) (ee)1/( f f )3( g g ) (hh) (ii))( j j )-( k k ) y^(( l l ) (mm) (nn)1/( o o )3( p p ) (qq) (rr))( s s )=c (tt)` (uu)

Options

A. `x^(2//3)-y^(2//3)=c`
B. `x^(4//3)-y^(4//3)=c`
C. `x^(4//3)+y^(4//3)=c`
D. `x^(1/3)-y^(1//3)=c`

Explanation

To find the orthogonal trajectories of the family of curves given by the equation:

\[ x^{2/3} + y^{2/3} = a^{2/3} \]

where \( a \) is an arbitrary constant, we will follow these steps:

### Step 1: Differentiate the given equation
We start by differentiating the equation with respect to \( x \):

\[ \frac{d}{dx}(x^{2/3}) + \frac{d}{dx}(y^{2/3}) = \frac{d}{dx}(a^{2/3}) \]

Since \( a \) is a constant, its derivative is zero. Thus, we have:

\[ \frac{2}{3} x^{-1/3} + \frac{2}{3} y^{-1/3} \frac{dy}{dx} = 0 \]

### Step 2: Rearranging the equation
Rearranging the equation gives:

\[ \frac{2}{3} y^{-1/3} \frac{dy}{dx} = -\frac{2}{3} x^{-1/3} \]

Now, we can cancel \( \frac{2}{3} \) from both sides:

\[ y^{-1/3} \frac{dy}{dx} = -x^{-1/3} \]

### Step 3: Express \(\frac{dy}{dx}\)
Now, we can express \(\frac{dy}{dx}\):

\[ \frac{dy}{dx} = -\frac{y^{1/3}}{x^{1/3}} \]

### Step 4: Finding the orthogonal trajectories
The slopes of the orthogonal trajectories are the negative reciprocals of the slopes of the original curves. Thus, we have:

\[ \frac{dy}{dx} = \frac{x^{1/3}}{y^{1/3}} \]

### Step 5: Separate variables
We can separate the variables:

\[ y^{1/3} dy = x^{1/3} dx \]

### Step 6: Integrate both sides
Now, we integrate both sides:

\[ \int y^{1/3} dy = \int x^{1/3} dx \]

Using the integration formula \(\int x^n dx = \frac{x^{n+1}}{n+1} + C\):

\[ \frac{y^{4/3}}{4/3} = \frac{x^{4/3}}{4/3} + C \]

This simplifies to:

\[ y^{4/3} = x^{4/3} + C \]

### Step 7: Rearranging the equation
Rearranging gives us the final form of the orthogonal trajectories:

\[ x^{4/3} - y^{4/3} = C \]

### Conclusion
Thus, the orthogonal trajectories of the given family of curves are represented by the equation:

\[ x^{4/3} - y^{4/3} = C \]