The normal at the.point (1, 1) on the curve `2y = 3 - x^2` is

Options

A. `x+y=0`
B. `x+y+1=0`
C. `x-y+1=0`
D. `x-y=0`

Answer

To find the normal at the point (1, 1) on the curve given by the equation \(2y = 3 - x^2\), we will follow these steps:

### Step 1: Rewrite the Curve Equation
First, we rewrite the curve equation in a more standard form:
\[ y = \frac{3 - x^2}{2} \]

### Step 2: Differentiate the Curve
Next, we differentiate the equation with respect to \(x\) to find the slope of the tangent line:
\[ \frac{dy}{dx} = \frac{d}{dx}\left(\frac{3 - x^2}{2}\right) = \frac{-2x}{2} = -x \]

### Step 3: Find the Slope of the Tangent at (1, 1)
Now we will find the slope of the tangent line at the point (1, 1):
\[ \text{slope of tangent} = \frac{dy}{dx} \bigg|_{x=1} = -1 \]

### Step 4: Find the Slope of the Normal
The slope of the normal line is the negative reciprocal of the slope of the tangent:
\[ \text{slope of normal} = -\frac{1}{\text{slope of tangent}} = -\frac{1}{-1} = 1 \]

### Step 5: Write the Equation of the Normal Line
Using the point-slope form of the line equation, we can write the equation of the normal line that passes through the point (1, 1):
\[ y - y_1 = m(x - x_1) \]
Substituting \(m = 1\), \(x_1 = 1\), and \(y_1 = 1\):
\[ y - 1 = 1(x - 1) \]
This simplifies to:
\[ y - 1 = x - 1 \]
Rearranging gives:
\[ x - y = 0 \quad \text{or} \quad x - y = 0 \]

### Step 6: Final Equation of the Normal
Thus, the equation of the normal at the point (1, 1) is:
\[ x - y = 0 \]