Equation of Normal to Curve \(y = 4x - 3x^2\) at Point \((2, -4)\)

To find the equation of the normal to a curve at a given point, we first need to determine the slope of the tangent to the curve at that point. The normal is then perpendicular to the tangent, allowing us to find its slope and subsequently its equation using the point-slope form.

1. Curve and Point Identification

• The given curve is \(y = 4x - 3x^2\).
• The specific point on the curve is \((x_1, y_1) = (2, -4)\).

2. Deriving Slope of Tangent

The slope of the tangent to the curve at any point \(x\) is given by the first derivative of the curve's equation with respect to \(x\), i.e., \(\frac{dy}{dx}\).

Let's differentiate \(y = 4x - 3x^2\) with respect to \(x\):

\[ \frac{dy}{dx} = \frac{d}{dx}(4x - 3x^2) \] \[ \frac{dy}{dx} = 4 \cdot \frac{d}{dx}(x) - 3 \cdot \frac{d}{dx}(x^2) \] \[ \frac{dy}{dx} = 4(1) - 3(2x) \] \[ \frac{dy}{dx} = 4 - 6x \]

3. Slope of Tangent at Specific Point \((2, -4)\)

Now, we need to find the numerical value of the slope of the tangent at the given point \((2, -4)\). We substitute \(x = 2\) into the derivative \(\frac{dy}{dx}\):

Slope of tangent, \(m_t = \left(\frac{dy}{dx}\right)_{x=2}\) \[ m_t = 4 - 6(2) \] \[ m_t = 4 - 12 \] \[ m_t = -8 \]

So, the slope of the tangent to the curve \(y = 4x - 3x^2\) at \((2, -4)\) is \(-8\).

4. Calculating Slope of Normal

The normal to a curve at a point is a line perpendicular to the tangent at that same point. If \(m_t\) is the slope of the tangent and \(m_n\) is the slope of the normal, then their product is \(-1\), provided neither slope is zero or undefined.

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

Using the calculated tangent slope \(m_t = -8\):

\[ m_n = -\frac{1}{-8} \] \[ m_n = \frac{1}{8} \]

Thus, the slope of the normal to the curve at \((2, -4)\) is \(\frac{1}{8}\).

5. Formulating Equation of Normal

We have the slope of the normal, \(m_n = \frac{1}{8}\), and the point it passes through, \((x_1, y_1) = (2, -4)\). We can use the point-slope form of a linear equation, which is \(y - y_1 = m(x - x_1)\).

Substitute the values:

\[ y - (-4) = \frac{1}{8}(x - 2) \] \[ y + 4 = \frac{1}{8}(x - 2) \]

To eliminate the fraction, multiply both sides of the equation by 8:

\[ 8(y + 4) = 8 \cdot \frac{1}{8}(x - 2) \] \[ 8y + 32 = x - 2 \]

Now, rearrange the equation into the general form \(Ax + By + C = 0\):

\[ 0 = x - 8y - 2 - 32 \] \[ x - 8y - 34 = 0 \]

6. Final Equation of Normal

The equation of the normal to the curve \(y = 4x - 3x^2\) at the point \((2, -4)\) is \(x - 8y - 34 = 0\).

Comparing this result with the given options, we find that it matches one of the choices.