Here's a rewritten solution designed to be more elaborate, clear, and educational for a JEE aspirant.

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Key Concept: The Fundamental Definition of a Parabola

A parabola is defined as the locus of a point $P(x,y)$ such that its distance from a fixed point, called the focus ($F$), is always equal to its perpendicular distance from a fixed straight line, called the directrix ($L$). Mathematically, for any point $P$ on the parabola, we have $PF = PM$, where $M$ is the foot of the perpendicular from $P$ to the directrix. This fundamental property is the most robust way to derive the equation of a parabola or to solve problems involving its definition, especially when the axis is not parallel to the coordinate axes.

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Step 1: Determine the Coordinates of the Vertex and Focus

Goal: Our first objective is to precisely locate the vertex $V$ and the focus $F$ of the parabola using the given geometric information.

Given Information:

• The axis of the parabola is the line $y=x$. This implies that both the vertex and focus lie on this line.
• Both the vertex $V$ and focus $F$ are situated in the first quadrant. This is a critical piece of information that helps us uniquely determine their coordinates.
• The distance of the vertex $V$ from the origin $O(0,0)$ is $OV = \sqrt{2}$ units.
• The distance of the focus $F$ from the origin $O(0,0)$ is $OF = 2\sqrt{2}$ units.

1.1 Finding the Coordinates of the Vertex $V$:

• Reasoning: Since $V$ lies on the line $y=x$, its coordinates must be of the form $(v, v)$. Furthermore, as $V$ is in the first quadrant, we know that $v$ must be positive ($v > 0$).
• Calculation: We use the distance formula to find the distance between the origin $O(0,0)$ and the vertex $V(v,v)$: $OV = \sqrt{(v-0)^2 + (v-0)^2} = \sqrt{v^2 + v^2} = \sqrt{2v^2}$ Since $v > 0$, we can simplify $\sqrt{2v^2}$ to $v\sqrt{2}$.
• Equating with the given distance: We are given that $OV = \sqrt{2}$. $v\sqrt{2} = \sqrt{2}$ Dividing both sides by $\sqrt{2}$, we get: $v = 1$
• Result: Therefore, the coordinates of the Vertex are $V=(1,1)$.

1.2 Finding the Coordinates of the Focus $F$:

• Reasoning: Similar to the vertex, the focus $F$ also lies on the axis $y=x$, so its coordinates are of the form $(f, f)$. Since $F$ is also in the first quadrant, $f$ must be positive ($f > 0$).
• Calculation: Using the distance formula for $F(f,f)$ from the origin $O(0,0)$: $OF = \sqrt{(f-0)^2 + (f-0)^2} = \sqrt{f^2 + f^2} = \sqrt{2f^2}$ Since $f > 0$, this simplifies to $f\sqrt{2}$.
• Equating with the given distance: We are given that $OF = 2\sqrt{2}$. $f\sqrt{2} = 2\sqrt{2}$ Dividing both sides by $\sqrt{2}$, we get: $f = 2$
• Result: Thus, the coordinates of the Focus are $F=(2,2)$.

Tip: Always leverage quadrant information to resolve ambiguities that might arise from square roots (e.g., $v=\pm 1$). It helps in selecting the correct coordinates. Also, notice that the focus $F(2,2)$ is further from the origin than the vertex $V(1,1)$, which is consistent with the standard arrangement of a parabola where the focus lies "inside" the curve, beyond the vertex.

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Step 2: Determine the Equation of the Directrix

Goal: Now that we have the vertex and focus, our next step is to find the equation of the directrix, which is essential for applying the parabola's definition.

Key Properties of a Parabola related to Directrix:

1. The vertex $V$ is the midpoint of the focus $F$ and the foot of the directrix on the axis of the parabola. Let's denote this foot as $M'$.
2. The directrix is a line perpendicular to the axis of the parabola.

2.1 Finding the Foot of the Directrix ($M'$):

• Reasoning: We know the vertex $V(1,1)$ is the midpoint of the focus $F(2,2)$ and the foot of the directrix $M'(x_d, y_d)$. We can use the midpoint formula to find the coordinates of $M'$. The midpoint formula states that if $(x_1, y_1)$ and $(x_2, y_2)$ are the endpoints of a segment, its midpoint is $\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$.
• Calculation: Let $F=(x_1, y_1)=(2,2)$ and $M'=(x_d, y_d)$. The midpoint is $V=(1,1)$. For the x-coordinate: $1 = \frac{2+x_d}{2} \Rightarrow 2 = 2+x_d \Rightarrow x_d = 0$ For the y-coordinate: $1 = \frac{2+y_d}{2} \Rightarrow 2 = 2+y_d \Rightarrow y_d = 0$
• Result: Thus, the foot of the directrix on the axis is $M'=(0,0)$, which is the origin.

2.2 Finding the Equation of the Directrix:

• Reasoning: We know the directrix passes through $M'(0,0)$ and is perpendicular to the axis of the parabola, which is the line $y=x$.
• Slope of the Axis: The equation of the axis is $y=x$, or $x-y=0$. Its slope, $m_{axis}$, is $1$.
• Slope of the Directrix: Since the directrix is perpendicular to the axis, the product of their slopes must be $-1$. $m_{directrix} \cdot m_{axis} = -1$ $m_{directrix} \cdot 1 = -1 \Rightarrow m_{directrix} = -1$
• Equation of the Directrix: Now we have a point $M'(0,0)$ on the directrix and its slope $m_{directrix} = -1$. Using the point-slope form of a linear equation, $y-y_1 = m(x-x_1)$: $y-0 = -1(x-0)$ $y = -x$ Rearranging this into the general form $Ax+By+C=0$: $x+y=0$
• Result: Thus, the Equation of the Directrix is $x+y=0$.

Tip: Visualizing these geometric relationships can greatly simplify the process. Imagine the axis as a mirror, the vertex as the point on the mirror, and the focus and directrix foot as equidistant points from the mirror.

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Step 3: Apply the Definition of Parabola ($PF=PM$)

Goal: With the focus and directrix determined, we can now use the definition of a parabola to form an equation involving the given point $P(1,k)$ and solve for $k$.

Known Information for this step:

• A point on the parabola: $P(1,k)$
• Focus: $F(2,2)$
• Directrix: $L: x+y=0$

3.1 Calculate $PF$ (Distance from Point P to Focus):

• Reasoning: We use the standard distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$, which is $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.
• Calculation: For $P(1,k)$ and $F(2,2)$: $PF = \sqrt{(1-2)^2 + (k-2)^2}$ $PF = \sqrt{(-1)^2 + (k-2)^2}$ $PF = \sqrt{1 + (k-2)^2}$

3.2 Calculate $PM$ (Perpendicular Distance from Point P to Directrix):

• Reasoning: We use the formula for the perpendicular distance from a point $(x_0, y_0)$ to a line $Ax+By+C=0$, which is $PM = \frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}$.
• Calculation: For $P(1,k)$ and the directrix $x+y=0$ (where $A=1, B=1, C=0$): $PM = \frac{|1(1) + 1(k) + 0|}{\sqrt{1^2 + 1^2}}$ $PM = \frac{|1+k|}{\sqrt{2}}$

3.3 Equate $PF$ and $PM$ and Solve for $k$:

• Reasoning: According to the fundamental definition of a parabola, the distance from any point on the parabola to the focus must be equal to its perpendicular distance to the directrix. Therefore, $PF = PM$.
• Calculation: $\sqrt{1 + (k-2)^2} = \frac{|1+k|}{\sqrt{2}}$ To eliminate the square roots and the absolute value, we square both sides of the equation. Squaring removes the absolute value since $(|A|)^2 = A^2$. $\left(\sqrt{1 + (k-2)^2}\right)^2 = \left(\frac{|1+k|}{\sqrt{2}}\right)^2$ $1 + (k-2)^2 = \frac{(1+k)^2}{2}$ Now, expand the squared terms: $1 + (k^2 - 4k + 4) = \frac{1 + 2k + k^2}{2}$ Combine terms on the left side: $k^2 - 4k + 5 = \frac{k^2 + 2k + 1}{2}$ Multiply both sides by 2 to clear the denominator: $2(k^2 - 4k + 5) = k^2 + 2k + 1$ $2k^2 - 8k + 10 = k^2 + 2k + 1$ Rearrange all terms to one side to form a standard quadratic equation $ak^2+bk+c=0$: $2k^2 - k^2 - 8k - 2k + 10 - 1 = 0$ $k^2 - 10k + 9 = 0$ Factor the quadratic equation. We look for two numbers that multiply to 9 and add up to -10 (which are -1 and -9): $(k-1)(k-9) = 0$ This yields two possible values for $k$: $k=1 \quad \text{or} \quad k=9$

Common Mistake to Avoid: Forgetting to square the denominator ($\sqrt{2}$) when squaring the right side, or incorrectly expanding $(k-2)^2$ or $(1+k)^2$. Always be meticulous with algebraic expansion.

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Step 4: Conclusion and Final Answer

Goal: Select the correct value of $k$ from the given options.

• The possible values of $k$ we found are 1 and 9.
• Let's consider the meaning of these values. If $k=1$, the point is $(1,1)$, which we identified as the vertex $V$. The vertex is, by definition, a point on the parabola, so $k=1$ is a valid solution.
• The given options are (A) 8, (B) 3, (C) 9, (D) 4.
• Comparing our solutions with the options, we see that $k=9$ is present as option (C).

Therefore, a possible value of $k$ is 9.

Final Answer: The final answer is $\boxed{\text{9}}$

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Summary and Key Takeaways:

This problem beautifully demonstrates how to solve complex parabola questions by strictly adhering to the fundamental definition ($PF=PM$).

1. Extract Key Information: Carefully read the problem to identify the axis, quadrant information, and distances.
2. Locate Focus and Vertex: Use the axis equation and distance formulas to pinpoint $V$ and $F$. Quadrant information is crucial for uniqueness.
3. Determine Directrix: Understand the geometric relationship: $V$ is the midpoint of $F