Key Concepts and Formulas

• Slope of Tangent and Normal: If $y = f(x)$, the slope of the tangent at a point $(x_1, y_1)$ is given by $\frac{dy}{dx}\Big|_{(x_1, y_1)}$. The slope of the normal is the negative reciprocal of the tangent's slope: $m_n = -\frac{1}{m_t}$.
• Equation of a Line: The equation of a line passing through $(x_1, y_1)$ with slope $m$ is $y - y_1 = m(x - x_1)$.
• Area of a Triangle: The area of a triangle with vertices $(0,0)$, $(x_1, y_1)$, and $(x_2, y_2)$ is given by $\frac{1}{2} |x_1y_2 - x_2y_1|$.

Step-by-Step Solution

Step 1: Find the derivative of the curve.

We are given the curve $y = 2x^2 + x + 2$. We need to find the derivative $\frac{dy}{dx}$ to determine the slope of the tangent. $\frac{dy}{dx} = \frac{d}{dx}(2x^2 + x + 2) = 4x + 1$

Step 2: Find the slope of the normal at point P.

Let $P(x_1, y_1)$ be a point on the curve. The slope of the tangent at $P$ is $m_t = 4x_1 + 1$. The slope of the normal at $P$ is the negative reciprocal of the tangent's slope: $m_n = -\frac{1}{4x_1 + 1}$

Step 3: Form the equation of the normal line l.

The equation of the normal line l passing through $P(x_1, y_1)$ with slope $m_n$ is: $y - y_1 = m_n(x - x_1)$ $y - y_1 = -\frac{1}{4x_1 + 1}(x - x_1)$

Step 4: Use point Q to determine the coordinates of P.

We are given that $Q(6, 4)$ lies on the normal line l. Substitute $x = 6$ and $y = 4$ into the equation of the normal line: $4 - y_1 = -\frac{1}{4x_1 + 1}(6 - x_1)$ Since $P(x_1, y_1)$ lies on the curve $y = 2x^2 + x + 2$, we have $y_1 = 2x_1^2 + x_1 + 2$. Substitute this into the equation above: $4 - (2x_1^2 + x_1 + 2) = -\frac{1}{4x_1 + 1}(6 - x_1)$ $2 - 2x_1^2 - x_1 = -\frac{6 - x_1}{4x_1 + 1}$ Multiply both sides by $(4x_1 + 1)$: $(2 - 2x_1^2 - x_1)(4x_1 + 1) = -(6 - x_1)$ $8x_1 + 2 - 8x_1^3 - 2x_1^2 - 4x_1^2 - x_1 = -6 + x_1$ $-8x_1^3 - 6x_1^2 + 7x_1 + 2 = -6 + x_1$ $8x_1^3 + 6x_1^2 - 6x_1 - 8 = 0$ Divide by 2: $4x_1^3 + 3x_1^2 - 3x_1 - 4 = 0$ By inspection, $x_1 = 1$ is a root: $4(1)^3 + 3(1)^2 - 3(1) - 4 = 4 + 3 - 3 - 4 = 0$ So, $x_1 = 1$. Now find $y_1$: $y_1 = 2(1)^2 + 1 + 2 = 2 + 1 + 2 = 5$ Thus, $P = (1, 5)$.

Step 5: Calculate the area of triangle OPQ.

We have $O(0, 0)$, $P(1, 5)$, and $Q(6, 4)$. The area of triangle $OPQ$ is: $\text{Area} = \frac{1}{2} |(1)(4) - (6)(5)| = \frac{1}{2} |4 - 30| = \frac{1}{2} |-26| = \frac{1}{2}(26) = 13$

Common Mistakes & Tips

• Sign Errors: Pay close attention to signs when calculating the slope of the normal and when expanding expressions.
• Algebraic Simplification: Double-check your algebraic manipulations to avoid errors, especially when dealing with polynomial equations.
• Root Finding: When solving cubic equations, try simple integer roots first (like $\pm 1, \pm 2$) before resorting to more complex methods.

Summary

We found the equation of the normal to the curve at point P, used the fact that point Q lies on the normal to find the coordinates of P, and then calculated the area of triangle OPQ using the determinant formula. The area of triangle OPQ is 13 square units.

The final answer is \boxed{13}.