If the line `x+y=0` touches the curve `2y^(2)=ax^(2)+b` at (1, -1), find a and b.

Answer

To solve the problem, we need to find the values of \( a \) and \( b \) such that the line \( x + y = 0 \) touches the curve \( 2y^2 = ax^2 + b \) at the point \( (1, -1) \).

### Step 1: Substitute the point into the curve equation
Since the point \( (1, -1) \) lies on the curve, we substitute \( x = 1 \) and \( y = -1 \) into the curve equation:

\[ 2(-1)^2 = a(1)^2 + b \]

This simplifies to:

\[ 2 = a + b \]

We will call this **Equation 1**.

### Step 2: Find the slope of the line
The line \( x + y = 0 \) can be rewritten as \( y = -x \). The slope of this line is:

\[ \text{slope of the line} = -1 \]

### Step 3: Differentiate the curve equation
Next, we differentiate the curve equation \( 2y^2 = ax^2 + b \) with respect to \( x \):

Using implicit differentiation:

\[ \frac{d}{dx}(2y^2) = \frac{d}{dx}(ax^2 + b) \]

This gives us:

\[ 4y \frac{dy}{dx} = 2ax \]

Now, solving for \( \frac{dy}{dx} \):

\[ \frac{dy}{dx} = \frac{2ax}{4y} = \frac{ax}{2y} \]

### Step 4: Substitute the point into the derivative
Now, we substitute \( x = 1 \) and \( y = -1 \) into the derivative to find the slope of the curve at that point:

\[ \frac{dy}{dx} = \frac{a(1)}{2(-1)} = -\frac{a}{2} \]

### Step 5: Set the slopes equal
Since the line touches the curve at this point, the slopes must be equal:

\[ -\frac{a}{2} = -1 \]

This simplifies to:

\[ \frac{a}{2} = 1 \implies a = 2 \]

We will call this **Equation 2**.

### Step 6: Substitute \( a \) back into Equation 1
Now that we have \( a = 2 \), we substitute this value back into Equation 1:

\[ 2 = 2 + b \]

Solving for \( b \):

\[ b = 2 - 2 = 0 \]

### Final Answer
Thus, the values of \( a \) and \( b \) are:

\[ a = 2, \quad b = 0 \]