If ` x^2 +y^2 =t -(1)/(t) , x^4 +y^4 =t^2 +(1)/(t^2),` prove that ` x^3 y (dy)/(dx)=1`

Answer

To prove that \( x^3 y \frac{dy}{dx} = 1 \), given the equations:

1. \( x^2 + y^2 = t - \frac{1}{t} \)
2. \( x^4 + y^4 = t^2 + \frac{1}{t^2} \)

we will follow these steps:

### Step 1: Square the first equation
We start with the first equation:

\[ x^2 + y^2 = t - \frac{1}{t} \]

Squaring both sides gives:

\[ (x^2 + y^2)^2 = \left(t - \frac{1}{t}\right)^2 \]

### Step 2: Expand both sides
Expanding both sides, we have:

\[ x^4 + 2x^2y^2 + y^4 = t^2 - 2 + \frac{1}{t^2} \]

### Step 3: Substitute the second equation
From the second equation, we know:

\[ x^4 + y^4 = t^2 + \frac{1}{t^2} \]

Substituting this into our expanded equation gives:

\[ t^2 + \frac{1}{t^2} + 2x^2y^2 = t^2 - 2 + \frac{1}{t^2} \]

### Step 4: Simplify the equation
Cancelling \( t^2 + \frac{1}{t^2} \) from both sides results in:

\[ 2x^2y^2 = -2 \]

Dividing both sides by 2 gives:

\[ x^2y^2 = -1 \]

### Step 5: Differentiate the equation
Now, we need to differentiate \( x^2 + y^2 = t - \frac{1}{t} \) with respect to \( x \):

Using implicit differentiation:

\[ \frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}\left(t - \frac{1}{t}\right) \]

This gives:

\[ 2x + 2y \frac{dy}{dx} = \frac{dt}{dx} + \frac{1}{t^2} \frac{dt}{dx} \]

Factoring out \( \frac{dt}{dx} \):

\[ 2x + 2y \frac{dy}{dx} = \left(1 + \frac{1}{t^2}\right) \frac{dt}{dx} \]

### Step 6: Solve for \( \frac{dy}{dx} \)
Rearranging gives:

\[ 2y \frac{dy}{dx} = \left(1 + \frac{1}{t^2}\right) \frac{dt}{dx} - 2x \]

Thus,

\[ \frac{dy}{dx} = \frac{\left(1 + \frac{1}{t^2}\right) \frac{dt}{dx} - 2x}{2y} \]

### Step 7: Substitute \( \frac{dy}{dx} \) into \( x^3 y \frac{dy}{dx} \)
Now we multiply both sides by \( x^3 y \):

\[ x^3 y \frac{dy}{dx} = x^3 y \cdot \frac{\left(1 + \frac{1}{t^2}\right) \frac{dt}{dx} - 2x}{2y} \]

This simplifies to:

\[ x^3 \cdot \frac{1 + \frac{1}{t^2}}{2} \frac{dt}{dx} - x^4 \]

### Step 8: Prove that \( x^3 y \frac{dy}{dx} = 1 \)
Given \( x^2 y^2 = -1 \), we can conclude that:

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

Thus, we have proved that:

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