Finding the Equation of the Normal to a Curve

We are asked to find the equation of the normal line to the curve defined by the equation $2y + x^2 = 3$ at a specific point $(1, 1)$. To find the equation of the normal line, we first need to determine its slope. The slope of the normal is related to the slope of the tangent line at the same point.

Calculating the Slope of the Tangent

The slope of the tangent line at any point on the curve is given by the derivative $\frac{dy}{dx}$. We can find this by differentiating the equation of the curve with respect to $x$.

Given curve equation:

$$2y + x^2 = 3$$

Differentiate both sides with respect to $x$:

$$\frac{d}{dx}(2y) + \frac{d}{dx}(x^2) = \frac{d}{dx}(3)$$

Using the chain rule for $\frac{d}{dx}(2y)$ and the power rule for $\frac{d}{dx}(x^2)$ and the derivative of a constant:

$$2\frac{dy}{dx} + 2x = 0$$

Now, we solve for $\frac{dy}{dx}$, which represents the slope of the tangent line, $m_t$:

$$2\frac{dy}{dx} = -2x$$

$$\frac{dy}{dx} = -x$$

So, the slope of the tangent at any point $(x, y)$ on the curve is $-x$.

Determining the Slope of the Tangent at (1, 1)

We need the slope of the tangent specifically at the point $(1, 1)$. Substitute $x=1$ into the expression for $\frac{dy}{dx}$:

$$m_t = \left(\frac{dy}{dx}\right)_{(1, 1)} = -(1) = -1$$

The slope of the tangent at the point $(1, 1)$ is $-1$.

Finding the Slope of the Normal

The normal line is perpendicular to the tangent line at the point of tangency. The product of the slopes of two perpendicular lines (neither of which is vertical) is $-1$. If $m_t$ is the slope of the tangent and $m_n$ is the slope of the normal, then:

$$m_n \cdot m_t = -1$$

Using the slope of the tangent $m_t = -1$ at $(1, 1)$:

$$m_n \cdot (-1) = -1$$

$$m_n = \frac{-1}{-1} = 1$$

The slope of the normal at the point $(1, 1)$ is $1$.

Writing the Equation of the Normal Line

We have the slope of the normal line, $m_n = 1$, and a point it passes through, $(1, 1)$. We can use the point-slope form of a linear equation, which is $y - y_1 = m(x - x_1)$, where $(x_1, y_1)$ is the point and $m$ is the slope.

Substitute the point $(1, 1)$ and the slope $m_n = 1$ into the point-slope form:

$$y - 1 = 1(x - 1)$$

Simplify the equation:

$$y - 1 = x - 1$$

Rearrange the terms to get the equation in the form $Ax + By + C = 0$ or similar:

$$0 = x - y$$

Or:

$$x - y = 0$$

This is the equation of the normal at the point $(1, 1)$ on the curve $2y + x^2 = 3$. This equation of the normal matches one of the given options.

Let's review the steps to find the equation of the normal:

• Start with the equation of the curve.
• Differentiate implicitly to find the slope of the tangent $\frac{dy}{dx}$.
• Evaluate the slope of the tangent at the given point $(1, 1)$.
• Use the relationship $m_n = -\frac{1}{m_t}$ to find the slope of the normal.
• Use the point-slope form $y - y_1 = m_n(x - x_1)$ with the given point and the normal slope to find the equation of the normal.

Following these steps gives the equation of the normal line as $x - y = 0$.