Finding the Normal Line Equation

We need to find the equation of the normal line to the curve defined by $2y + x^2 = 3$ at the specific point $(1, 1)$.

Step 1: Implicit Differentiation

First, differentiate the equation $2y + x^2 = 3$ implicitly with respect to $x$ to find the slope of the tangent line ($dy/dx$).

• Differentiating term by term: $ \frac{d}{dx}(2y) + \frac{d}{dx}(x^2) = \frac{d}{dx}(3) $
• Applying differentiation rules: $ 2\frac{dy}{dx} + 2x = 0 $

Step 2: Calculate Tangent Slope

Solve for $\frac{dy}{dx}$ and evaluate it at the point $(1, 1)$ to find the slope of the tangent line ($m_{\text{tangent}}$).

• Isolate $\frac{dy}{dx}$: $ 2\frac{dy}{dx} = -2x $ $ \frac{dy}{dx} = -x $
• Substitute the point $(1, 1)$: $ m_{\text{tangent}} = \frac{dy}{dx}\bigg|_{(1,1)} = -(1) = -1 $

Step 3: Determine Normal Slope

The normal line is perpendicular to the tangent line. The slope of the normal line ($m_{\text{normal}}$) is the negative reciprocal of the tangent slope.

• Calculate $m_{\text{normal}}$: $ m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}} = -\frac{1}{-1} = 1 $

Step 4: Equation of the Normal Line

Use the point-slope form of a linear equation, $y - y_1 = m(x - x_1)$, using the point $(1, 1)$ and the normal slope $m_{\text{normal}} = 1$.

• Substitute the values: $ y - 1 = 1(x - 1) $
• Simplify the equation: $ y - 1 = x - 1 $ $ y = x $
• Rearrange to match the options: $ x - y = 0 $

The equation of the normal line at the point $(1, 1)$ is $x - y = 0$. This corresponds to Option B.