1. Understanding the Parabola: Key Concepts and Standard Equations

A parabola is a fundamental conic section defined as the locus of all points in a plane that are equidistant from a fixed point (called the focus) and a fixed line (called the directrix).

For a parabola whose axis lies along the x-axis, its standard equation with the vertex at $(h, k)$ is given by: $(y-k)^2 = 4a(x-h)$ Here:

• $(h, k)$ represents the coordinates of the vertex of the parabola.
• The term '$a$' is the focal length, which is the distance between the vertex and the focus. It is also the distance between the vertex and the directrix.
• The sign of '$a$' determines the opening direction of the parabola:
  • If $a > 0$, the parabola opens to the right. The focus is at $(h+a, k)$ and the directrix is $x = h-a$.
  • If $a < 0$, the parabola opens to the left. The focus is at $(h+a, k)$ and the directrix is $x = h-a$. (Note: sometimes $a$ is taken as a positive distance and the $4a$ term is negative to indicate opening left). In this solution, we will assume $a$ can be positive or negative, directly indicating direction.

2. Determining the Vertex of the Parabola

The problem states that the axis of the parabola lies along the x-axis. This means the y-coordinate of the vertex will be 0, i.e., $k=0$. We are given that its vertex is at a distance of 2 units from the origin, on the positive x-axis.

• Why this step? The vertex is a crucial point for defining the parabola's equation. Its coordinates $(h, k)$ are directly used in the standard form.
• Since the vertex is on the positive x-axis at a distance of 2 from the origin, its coordinates are $(2, 0)$.
• Therefore, we have $h = 2$ and $k = 0$.

3. Determining the Focus and Focal Length 'a'

The problem states that the focus is at a distance of 4 units from the origin, on the positive x-axis.

• Why this step? The focus, along with the vertex, allows us to determine the focal length '$a$', which is essential for writing the complete equation of the parabola.
• Since the focus is on the positive x-axis at a distance of 4 from the origin, its coordinates are $(4, 0)$.
• For a parabola opening along the x-axis with vertex $(h, k)$, the focus is located at $(h+a, k)$ if it opens to the right, or $(h-a, k)$ if it opens to the left.
• We have the vertex $V=(2, 0)$ and the focus $F=(4, 0)$.
• Since the x-coordinate of the focus (4) is greater than the x-coordinate of the vertex (2), the focus is to the right of the vertex. This implies the parabola opens to the right, so we use the form $(h+a, k)$ for the focus.
• Equating the coordinates of the focus: $(h+a, k) = (4, 0)$
• Substitute $h=2$ and $k=0$: $(2+a, 0) = (4, 0)$
• Comparing the x-coordinates: $2+a = 4$ $a = 4 - 2$ $a = 2$
• Tip: Always visualize the positions of the vertex and focus. If the focus is to the right of the vertex (for x-axis aligned parabola) or above the vertex (for y-axis aligned parabola), 'a' will be positive. If it's to the left or below, 'a' will be negative (or you can keep 'a' positive and adjust the sign in $4a(x-h)$ to $4(-a)(x-h)$).

4. Formulating the Equation of the Parabola

Now that we have $h=2$, $k=0$, and $a=2$, we can substitute these values into the standard equation of the parabola: $(y-k)^2 = 4a(x-h)$

• Why this step? This is the ultimate goal before checking the options. The equation defines all points that lie on the parabola.
• Substitute the values: $(y-0)^2 = 4(2)(x-2)$ $y^2 = 8(x-2)$ $y^2 = 8x - 16$ This is the equation of the given parabola.

5. Checking Which Point Does Not Lie on the Parabola

To determine which point does not lie on the parabola, we substitute the coordinates of each option into the derived equation $y^2 = 8x - 16$. If the equation holds true (LHS = RHS), the point lies on the parabola. If not, it doesn't.

• Why this step? The question asks which point does not lie on the parabola. This requires a systematic check of all given options against the derived equation.

(A) Check point (5, $2\sqrt{6}$): Substitute $x=5$ and $y=2\sqrt{6}$ into $y^2 = 8x - 16$: LHS: $(2\sqrt{6})^2 = 2^2 \times (\sqrt{6})^2 = 4 \times 6 = 24$ RHS: $8(5) - 16 = 40 - 16 = 24$ Since LHS = RHS ($24 = 24$), the point $(5, 2\sqrt{6})$ lies on the parabola.

(B) Check point (6, $4\sqrt{2}$): Substitute $x=6$ and $y=4\sqrt{2}$ into $y^2 = 8x - 16$: LHS: $(4\sqrt{2})^2 = 4^2 \times (\sqrt{2})^2 = 16 \times 2 = 32$ RHS: $8(6) - 16 = 48 - 16 = 32$ Since LHS = RHS ($32 = 32$), the point $(6, 4\sqrt{2})$ lies on the parabola.

(C) Check point (8, 6): Substitute $x=8$ and $y=6$ into $y^2 = 8x - 16$: LHS: $6^2 = 36$ RHS: $8(8) - 16 = 64 - 16 = 48$ Since LHS $\neq$ RHS ($36 \neq 48$), the point $(8, 6)$ does not lie on the parabola.

(D) Check point (4, -4): Substitute $x=4$ and $y=-4$ into $y^2 = 8x - 16$: LHS: $(-4)^2 = 16$ RHS: $8(4) - 16 = 32 - 16 = 16$ Since LHS = RHS ($16 = 16$), the point $(4, -4)$ lies on the parabola.

Conclusion: Based on our checks, the point $(8, 6)$ is the only one that does not satisfy the equation of the parabola.

Common Mistakes to Avoid:

• Incorrectly identifying $h, k, a$: Always carefully read the distances and directions (positive/negative x-axis).
• Confusing axis alignment: If the axis were along the y-axis, the standard equation would be $(x-h)^2 = 4a(y-k)$.
• Sign errors: Be meticulous with positive and negative signs, especially when calculating $a$ or substituting values.
• Algebraic errors: Squaring terms or multiplying incorrectly can lead to wrong conclusions.

Summary and Key Takeaway:

To solve this problem, we first leveraged the given information about the vertex and focus to determine the specific equation of the parabola. We identified the vertex $(h, k)$ as $(2,0)$ and used the focus $(4,0)$ to calculate the focal length $a=2$. Substituting these into the standard equation $(y-k)^2 = 4a(x-h)$, we obtained $y^2 = 8(x-2)$. Finally, we systematically checked each given option by substituting its coordinates into this equation. The point that did not satisfy the equation was the answer. This method highlights the importance of correctly identifying the parameters of the conic section and careful algebraic verification.

The final answer is $\boxed{\text{(C)}}$.