Slope of the normal to the curve :
`y^(2)=4x` at `(1, 2)` is :

Options

A. 1
B. `(1)/(2)`
C. 2
D. `-1`

Answer

To find the slope of the normal to the curve \( y^2 = 4x \) at the point \( (1, 2) \), we will follow these steps:

### Step 1: Differentiate the equation of the curve
We start with the equation of the curve:
\[ y^2 = 4x \]
We differentiate both sides with respect to \( x \):
\[ \frac{d}{dx}(y^2) = \frac{d}{dx}(4x) \]
Using the chain rule on the left side, we get:
\[ 2y \frac{dy}{dx} = 4 \]

### Step 2: Solve for \(\frac{dy}{dx}\)
Now, we can isolate \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{4}{2y} = \frac{2}{y} \]

### Step 3: Substitute the point into the derivative
Next, we substitute the point \( (1, 2) \) into the derivative to find the slope of the tangent line at that point:
\[ \frac{dy}{dx} \bigg|_{(1, 2)} = \frac{2}{2} = 1 \]
So, the slope of the tangent line at the point \( (1, 2) \) is \( 1 \).

### Step 4: Find the slope of the normal
The slope of the normal line is the negative reciprocal of the slope of the tangent line. Therefore, we calculate:
\[ \text{slope of normal} = -\frac{1}{\text{slope of tangent}} = -\frac{1}{1} = -1 \]

### Conclusion
Thus, the slope of the normal to the curve \( y^2 = 4x \) at the point \( (1, 2) \) is:
\[ \boxed{-1} \]

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