Slope of the normal to the curve `y=2x^(2)-1` at `(1, 1)` is …………… .

Answer

To find the slope of the normal to the curve \( y = 2x^2 - 1 \) at the point \( (1, 1) \), we will follow these steps:

### Step 1: Differentiate the function
We start by differentiating the function \( y = 2x^2 - 1 \) with respect to \( x \) to find the slope of the tangent line.

\[ \frac{dy}{dx} = \frac{d}{dx}(2x^2 - 1) \]

Using the power rule, we differentiate:

\[ \frac{dy}{dx} = 2 \cdot 2x^{2-1} + 0 = 4x \]

### Step 2: Evaluate the derivative at the given point
Next, we evaluate the derivative at the point \( x = 1 \) to find the slope of the tangent line at that point.

\[ \frac{dy}{dx} \bigg|_{x=1} = 4(1) = 4 \]

### Step 3: Find the slope of the normal
The slope of the normal line is the negative reciprocal of the slope of the tangent line. If the slope of the tangent is \( m_t = 4 \), then the slope of the normal \( m_n \) is given by:

\[ m_n = -\frac{1}{m_t} = -\frac{1}{4} \]

### Conclusion
Thus, the slope of the normal to the curve \( y = 2x^2 - 1 \) at the point \( (1, 1) \) is

\[ \boxed{-\frac{1}{4}} \]

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