Key Concepts and Formulas

• Implicit Differentiation: Used to find the derivative $\frac{dy}{dx}$ when $y$ is not explicitly defined as a function of $x$. Remember to apply the chain rule when differentiating terms involving $y$.
• Equation of a Tangent Line: Given a point $(x_1, y_1)$ on a curve and the slope $m$ of the tangent at that point, the equation of the tangent line is $y - y_1 = m(x - x_1)$.
• Equation of a Normal Line: The normal line is perpendicular to the tangent line. Its slope is $-\frac{1}{m}$, where $m$ is the slope of the tangent. The equation of the normal line is $y - y_1 = -\frac{1}{m}(x - x_1)$.
• Area of a Triangle: Given the vertices $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$, the area of the triangle is given by $A = \frac{1}{2} |x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|$. Alternatively, if the triangle is formed by the x-axis and two lines, it can be found as $A = \frac{1}{2} |x_2 - x_1| |y|$ where $x_1$ and $x_2$ are the x-intercepts of the two lines and $y$ is the y-coordinate of the intersection point.

Step-by-Step Solution

Step 1: Implicit Differentiation

We need to find $\frac{dy}{dx}$ at the point $(-2, 3)$. Differentiate the given equation implicitly with respect to $x$: $4{x^3} - 3x{y^2} + 6{x^2} - 5xy - 8{y^2} + 9x + 14 = 0$ Differentiating each term: $12x^2 - 3(y^2 + x(2y\frac{dy}{dx})) + 12x - 5(y + x\frac{dy}{dx}) - 16y\frac{dy}{dx} + 9 = 0$ $12x^2 - 3y^2 - 6xy\frac{dy}{dx} + 12x - 5y - 5x\frac{dy}{dx} - 16y\frac{dy}{dx} + 9 = 0$ Now, group the terms with $\frac{dy}{dx}$: $\frac{dy}{dx}(-6xy - 5x - 16y) = -12x^2 + 3y^2 - 12x + 5y - 9$ $\frac{dy}{dx} = \frac{-12x^2 + 3y^2 - 12x + 5y - 9}{-6xy - 5x - 16y}$

Step 2: Find the slope at (-2, 3)

Substitute $x = -2$ and $y = 3$ into the expression for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{-12(-2)^2 + 3(3)^2 - 12(-2) + 5(3) - 9}{-6(-2)(3) - 5(-2) - 16(3)}$ $\frac{dy}{dx} = \frac{-12(4) + 3(9) + 24 + 15 - 9}{36 + 10 - 48}$ $\frac{dy}{dx} = \frac{-48 + 27 + 24 + 15 - 9}{46 - 48}$ $\frac{dy}{dx} = \frac{9}{-2} = -\frac{9}{2}$ So, the slope of the tangent at $(-2, 3)$ is $m_t = -\frac{9}{2}$.

Step 3: Find the equation of the tangent line

Using the point-slope form, the equation of the tangent line is: $y - 3 = -\frac{9}{2}(x + 2)$ $2(y - 3) = -9(x + 2)$ $2y - 6 = -9x - 18$ $9x + 2y + 12 = 0$

Step 4: Find the x-intercept of the tangent line

Set $y = 0$ to find the x-intercept: $9x + 12 = 0$ $9x = -12$ $x = -\frac{12}{9} = -\frac{4}{3}$ The x-intercept of the tangent line is $(-\frac{4}{3}, 0)$.

Step 5: Find the equation of the normal line

The slope of the normal line is the negative reciprocal of the tangent's slope: $m_n = -\frac{1}{m_t} = -\frac{1}{-\frac{9}{2}} = \frac{2}{9}$. The equation of the normal line is: $y - 3 = \frac{2}{9}(x + 2)$ $9(y - 3) = 2(x + 2)$ $9y - 27 = 2x + 4$ $2x - 9y + 31 = 0$

Step 6: Find the x-intercept of the normal line

Set $y = 0$ to find the x-intercept: $2x + 31 = 0$ $2x = -31$ $x = -\frac{31}{2}$ The x-intercept of the normal line is $(-\frac{31}{2}, 0)$.

Step 7: Calculate the area of the triangle

The vertices of the triangle are $(-4/3, 0)$, $(-31/2, 0)$, and $(-2, 3)$. The area of the triangle formed by the x-axis, the tangent, and the normal is given by: $A = \frac{1}{2} |x_2 - x_1| |y|$ where $x_1 = -\frac{31}{2}$, $x_2 = -\frac{4}{3}$, and $y = 3$. $A = \frac{1}{2} \left|-\frac{4}{3} - \left(-\frac{31}{2}\right)\right| (3)$ $A = \frac{3}{2} \left|-\frac{4}{3} + \frac{31}{2}\right|$ $A = \frac{3}{2} \left|\frac{-8 + 93}{6}\right|$ $A = \frac{3}{2} \left|\frac{85}{6}\right|$ $A = \frac{3}{2} \cdot \frac{85}{6} = \frac{85}{4}$

Step 8: Calculate 8A

$8A = 8 \cdot \frac{85}{4} = 2 \cdot 85 = 170$

Common Mistakes & Tips

• Sign Errors: Pay close attention to signs when differentiating and substituting values. A small sign error can lead to a completely wrong answer.
• Implicit Differentiation: Remember to apply the chain rule correctly when differentiating terms involving $y$.
• Fractions: Work carefully with fractions to avoid arithmetic errors.

Summary

We first found the derivative of the given equation using implicit differentiation. Then, we calculated the slope of the tangent and normal lines at the given point. Next, we determined the equations of the tangent and normal lines and found their x-intercepts. Finally, we used the x-intercepts and the y-coordinate of the intersection point of the tangent and normal to calculate the area of the triangle, which was then multiplied by 8. The final answer is 170.

Final Answer

The final answer is \boxed{170}.