This problem requires a comprehensive understanding of parabolas, including finding equations of tangents and normals, determining intersection points, and calculating the area of a triangle. We will break down the solution into several distinct steps, each building upon the previous one.

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1. Introduction and Problem Overview

We are given a parabola $S: y^2 = 2x$ and a specific point $P(2, 2)$ that lies on it. Our goal is to:

1. Determine the equation of the tangent line to the parabola at point $P$.
2. Find the point $Q$ where this tangent line intersects the x-axis.
3. Determine the equation of the normal line to the parabola at point $P$.
4. Find the point $R$ where this normal line intersects the parabola again (other than $P$).
5. Calculate the area of the triangle formed by the three points $P, Q,$ and $R$.

This problem serves as an excellent test of fundamental concepts in coordinate geometry, particularly those related to conic sections.

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2. Understanding the Parabola and Point P

Before performing any calculations, it's crucial to correctly identify the parameters of the given parabola and confirm the point $P$ lies on it.

• Key Concept: The standard form of a parabola symmetric about the x-axis is $y^2 = 4ax$.

• Applying the concept: Our given parabola is $y^2 = 2x$. By comparing $y^2 = 2x$ with $y^2 = 4ax$, we can determine the parameter $a$: $4a = 2 \implies a = \frac{2}{4} = \frac{1}{2}$ This value of $a = \frac{1}{2}$ will be used in various standard formulas for the parabola.

• Verification of Point P: It's good practice to ensure the given point $P(2, 2)$ actually lies on the parabola. Substitute $x=2$ and $y=2$ into the parabola's equation $y^2 = 2x$: Left Hand Side (LHS) $= y^2 = (2)^2 = 4$. Right Hand Side (RHS) $= 2x = 2(2) = 4$. Since LHS = RHS ($4=4$), the point $P(2, 2)$ indeed lies on the parabola $y^2 = 2x$. This confirms our starting point is valid.

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3. Step 1: Finding the Tangent Equation at P and Point Q

Our first task is to find the equation of the tangent line to the parabola at $P(2, 2)$ and then locate its intersection with the x-axis, which we'll call $Q$.

• Key Concept: The equation of the tangent to a parabola $y^2 = 4ax$ at a point $P(x_1, y_1)$ on the parabola is given by the formula $yy_1 = 2a(x+x_1)$. This formula is derived from the general method of finding tangents to conic sections or by implicit differentiation.

• Applying the formula: We have the parabola $y^2 = 2x$, so $2a = 1$. The point of tangency is $P(x_1, y_1) = (2, 2)$. Substitute these values into the tangent formula: $y(2) = 1(x+2)$ $2y = x+2$ Rearranging this into the standard form $Ax+By+C=0$ or slope-intercept form $y=mx+c$: $x - 2y + 2 = 0 \quad \text{or} \quad y = \frac{1}{2}x + 1$ This is the equation of the tangent line to the parabola at $P(2, 2)$.

• Finding Point Q: Point $Q$ is where the tangent line intersects the x-axis. Any point on the x-axis has a y-coordinate of 0. Why: We set $y=0$ to find the x-intercept because the x-axis is defined by $y=0$. Substitute $y=0$ into the tangent equation $2y = x+2$: $2(0) = x+2$ $0 = x+2$ $x = -2$ Therefore, the coordinates of point $Q$ are $(-2, 0)$.

• Tip (Alternative for Slope): You could also find the slope of the tangent by implicit differentiation. Differentiate $y^2 = 2x$ with respect to $x$: $2y \frac{dy}{dx} = 2 \implies \frac{dy}{dx} = \frac{1}{y}$. At $P(2, 2)$, the slope of the tangent $m_T = \frac{1}{2}$. Then use the point-slope form $y - y_1 = m_T(x - x_1)$: $y - 2 = \frac{1}{2}(x - 2)$ $2(y - 2) = x - 2$ $2y - 4 = x - 2$ $2y = x + 2$, which is the same tangent equation.

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4. Step 2: Finding the Normal Equation at P and Point R

Next, we need to find the equation of the normal line to the parabola at $P(2, 2)$ and then determine its second intersection point $R$ with the parabola.

• Key Concept (Slope of Normal): The normal to a curve at a point is a line perpendicular to the tangent at that point. If the slope of the tangent is $m_T$, then the slope of the normal, $m_N$, is its negative reciprocal: $m_N = -\frac{1}{m_T}$ (provided $m_T \neq 0$).

• Key Concept (Equation of a Line): The equation of a line passing through a point $(x_1, y_1)$ with slope $m$ is $y - y_1 = m(x - x_1)$.

• Calculating the slope of the tangent ($m_T$): From the previous step, we found the slope of the tangent at $P(2, 2)$ was $m_T = \frac{1}{2}$. (If you used $2y=x+2$, rewrite it as $y=\frac{1}{2}x+1$ to see the slope directly).

• Calculating the slope of the normal ($m_N$): Why: We need the slope of the normal to write its equation. $m_N = -\frac{1}{m_T} = -\frac{1}{1/2} = -2$

• Applying the normal formula: We have the point $P(x_1, y_1) = (2, 2)$ and the slope of the normal $m_N = -2$. Substitute these into the point-slope form: $y - 2 = -2(x - 2)$ $y - 2 = -2x + 4$ $y = -2x + 6$ This is the equation of the normal line to the parabola at $P(2, 2)$.

• Finding Point R (second intersection with parabola): Point $R$ is the other point where the normal line $y = -2x + 6$ intersects the parabola $S: y^2 = 2x$. We already know one intersection point is $P(2, 2)$. Why: To find intersection points, we solve the equations of the line and the curve simultaneously. Substitute the expression for $y$ from the normal's equation into the parabola's equation: $(-2x + 6)^2 = 2x$ Expand the left side (using $(A-B)^2 = A^2 - 2AB + B^2$): $(2x)^2 - 2(2x)(6) + 6^2 = 2x$ $4x^2 - 24x + 36 = 2x$ Rearrange the terms to form a standard quadratic equation: $4x^2 - 26x + 36 = 0$ Divide the entire equation by 2 to simplify the coefficients: $2x^2 - 13x + 18 = 0$ This quadratic equation gives the x-coordinates of the intersection points. We know that $x_P = 2$ is one root (since $P$ is an intersection point). Let the other root be $x_R$.

• Using Vieta's formulas: For a quadratic equation $Ax^2 + Bx + C = 0$, the sum of the roots is $-B/A$. Why: This is a quick way to find the second root if one root is already known, avoiding the need to factorize or use the quadratic formula directly. The sum of the roots is $x_P + x_R = -\frac{(-13)}{2} = \frac{13}{2}$. Since $x_P = 2$: $2 + x_R = \frac{13}{2}$ Solve for $x_R$: $x_R = \frac{13}{2} - 2 = \frac{13}{2} - \frac{4}{2} = \frac{9}{2}$ Now that we have $x_R = \frac{9}{2}$, substitute this value into the normal's equation $y = -2x + 6$ to find $y_R$: $y_R = -2\left(\frac{9}{2}\right) + 6$ $y_R = -9 + 6$ $y_R = -3$ Therefore, the coordinates of point $R$ are $\left(\frac{9}{2}, -3\right)$.

• Tip (Parametric Form for Normals): For a parabola $y^2 = 4ax$, a point $P$ can be represented parametrically as $(at^2, 2at)$. The normal at $P(at^2, 2at)$ intersects the parabola again at $R(at'^2, 2at')$ where $t' = -t - \frac{2}{t}$. This formula is extremely useful for quickly finding the second intersection point of a normal. For $y^2 = 2x$, we have $a = \frac{1}{2}$. For $P(2, 2)$, we have $2at = 2 \implies 2\left(\frac{1}{2}\right)t = 2 \implies t = 2$. Using the formula for $t'$: $t' = -t - \frac{2}{t} = -2 - \frac{2}{2} = -2 - 1 = -3$ Now find the coordinates of $R$ using $t' = -3$: $x_R = at'^2 = \frac{1}{2}(-3)^2 = \frac{1}{2}(9) = \frac{9}{2}$ $y_R = 2at' = 2\left(\frac{1}{2}\right)(-3) = -3$ This confirms our calculated coordinates for $R$ are $\left(\frac{9}{2}, -3\right)$, demonstrating the efficiency of the parametric approach.

• Common Mistake: After finding $x_R$, remember to substitute it back into the normal's equation (or parabola's equation, but normal's is usually simpler) to find $y_R$. Don't just assume a value or forget this step.

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5. Step 3: Calculating the Area of Triangle PQR

Now that we have the coordinates of all three vertices, we can calculate the area of triangle PQR. The vertices are:

• $P(x_1, y_1) = (2, 2)$
• $Q(x_2, y_2)