Key Concepts and Formulas

This problem is a classic application of coordinate geometry, combining the properties of parabolas with the section formula for internal division of a line segment. Mastering these concepts is fundamental for JEE preparation.

1. Parabola Definition and Vertex: A parabola is defined as the set of all points in a plane that are equidistant from a fixed point (the focus) and a fixed straight line (the directrix).

   • The standard form of a parabola opening upwards or downwards is $x^2 = 4ay$.
     • Its vertex is at the origin $(0,0)$.
     • Its axis of symmetry is the y-axis.
     • If $a > 0$, it opens upwards. If $a < 0$, it opens downwards.
   • The standard form of a parabola opening left or right is $y^2 = 4ax$.
     • Its vertex is also at the origin $(0,0)$.
     • Its axis of symmetry is the x-axis.
     • If $a > 0$, it opens to the right. If $a < 0$, it opens to the left.

2. Section Formula (Internal Division): This formula is used to find the coordinates of a point that divides a line segment joining two given points internally in a specific ratio.

   • If a point $P(x,y)$ divides the line segment joining $A(x_1, y_1)$ and $B(x_2, y_2)$ internally in the ratio $m:n$, then the coordinates of $P$ are given by: $P(x,y) = \left( \frac{nx_1 + mx_2}{m+n}, \frac{ny_1 + my_2}{m+n} \right)$
   • Important Note: In the ratio $m:n$, $m$ is associated with the segment closer to $B(x_2, y_2)$ and $n$ is associated with the segment closer to $A(x_1, y_1)$. A common mistake is to swap them.

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Understanding the Problem Setup

Let's carefully dissect the problem statement to understand what is given and what we need to find:

• Parabola: We are given the equation $x^2 = 8y$. This is a parabola.
• Vertex $O$: The problem states that $O$ is the vertex of this parabola. We need to identify its coordinates.
• Point $Q$: $Q$ is any arbitrary point on the parabola $x^2 = 8y$. This means its coordinates must satisfy the parabola's equation.
• Point $P$: $P$ divides the line segment $OQ$ internally in the ratio $1:3$. This is a crucial piece of information that links $P$ to $O$ and $Q$.
• Goal: We need to find the locus of $P$. This means we need to find an equation that describes all possible positions of point $P$ as point $Q$ moves along the given parabola. The locus will typically be another curve.

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Step-by-Step Solution

Step 1: Identify the coordinates of the vertex $O$ and define an arbitrary point $Q$.

• Identifying the Vertex $O$: The given equation of the parabola is $x^2 = 8y$. Comparing this to the standard form $x^2 = 4ay$, we can see that $4a = 8$, which implies $a = 2$. For any parabola of the form $x^2 = 4ay$ or $y^2 = 4ax$, the vertex is at the origin $(0,0)$. Therefore, the coordinates of the vertex $O$ are $\mathbf{(0,0)}$.

  • Why this step? The vertex $O$ is one of the endpoints of the line segment $OQ$. Knowing its exact coordinates is essential for applying the section formula.

• Defining Point $Q$: Let $Q$ be any arbitrary point on the parabola $x^2 = 8y$. We'll denote its coordinates as $(x_1, y_1)$. Since $Q$ lies on the parabola, its coordinates must satisfy the parabola's equation. So, we have the fundamental relationship: $x_1^2 = 8y_1 \quad \text{(Equation 1)}$

  • Why this step? We need to represent $Q$ generally because it can be any point on the parabola. Equation (1) is the constraint that $Q$ must satisfy, and it will be key to finding the locus of $P$.

• Defining Point $P$: We are trying to find the locus of point $P$. In locus problems, it is standard practice to denote the coordinates of the point whose locus is required as $(h,k)$. So, let $P = (h,k)$.

  • Why this step? By assigning $(h,k)$ to $P$, we establish the variables for which we need to find a relationship. Our ultimate goal is to find an equation involving $h$ and $k$.

Step 2: Apply the Section Formula to express the coordinates of $P$ in terms of $O$ and $Q$.

• We are given that point $P(h,k)$ divides the line segment $OQ$ internally in the ratio $1:3$.
• Let $O = (x_O, y_O) = (0,0)$.
• Let $Q = (x_Q, y_Q) = (x_1, y_1)$.
• The ratio of division is $m:n = 1:3$. (Here $m=1$ and $n=3$).
• Using the section formula: $h = \frac{n \cdot x_O + m \cdot x_Q}{m+n} \quad \text{and} \quad k = \frac{n \cdot y_O + m \cdot y_Q}{m+n}$ Substitute the coordinates of $O(0,0)$ and $Q(x_1, y_1)$, and the ratio $m=1, n=3$: $h = \frac{3(0) + 1(x_1)}{1+3} \implies h = \frac{x_1}{4}$ $k = \frac{3(0) + 1(y_1)}{1+3} \implies k = \frac{y_1}{4}$
  • Why this step? This establishes a direct algebraic link between the coordinates of $P(h,k)$ and the coordinates of $Q(x_1, y_1)$. Since $Q$ is constrained by the parabola's equation, this link will allow us to transfer that constraint to $P$.

Step 3: Express $x_1$ and $y_1$ (coordinates of $Q$) in terms of $h$ and $k$ (coordinates of $P$).

From the results of Step 2, we have: $h = \frac{x_1}{4} \implies x_1 = 4h \quad \text{(Equation 2)}$ $k = \frac{y_1}{4} \implies y_1 = 4k \quad \text{(Equation 3)}$

• Why this step? Our goal is to find an equation purely in terms of $h$ and $k$. To do this, we need to eliminate $x_1$ and $y_1$. By rearranging these equations, we prepare $x_1$ and $y_1$ for substitution into Equation 1, which is the defining equation for $Q$.

Step 4: Substitute $x_1$ and $y_1$ into the parabola's equation (Equation 1).

We know that $Q(x_1, y_1)$ lies on the parabola $x^2 = 8y$, so $x_1^2 = 8y_1$. Now, substitute Equation 2 and Equation 3 into Equation 1: $(4h)^2 = 8(4k)$ $16h^2 = 32k$ Now, simplify the equation by dividing both sides by 16: $h^2 = \frac{32k}{16}$ $h^2 = 2k \quad \text{(Equation 4)}$

• Why this step? This is the core of finding the locus. By substituting the expressions for $x_1$ and $y_1$ (which are in terms of $h$ and $k$) into the equation that $Q$ must satisfy, we effectively transfer the geometric constraint from $Q$ to $P$. The resulting equation is a relationship solely between $h$ and $k$, which defines the path of $P$.

Step 5: Replace $(h,k)$ with $(x,y)$ to represent the locus.

The equation $h^2 = 2k$ describes the relationship that the coordinates $(h,k)$ of point $P$ must satisfy. To represent this locus in the standard coordinate system, we simply replace $h$ with $x$ and $k$ with $y$. Therefore, the locus of point $P$ is: $\mathbf{x^2 = 2y}$

• Why this step? This is a convention for expressing the equation of a locus. $(h,k)$ are temporary variables used during the derivation to distinguish the point whose locus is being found from the general coordinates of other points. Once the relationship is established, we revert to $(x,y)$ for the final equation.

Step 6: Compare with the given options.

The derived equation for the locus of $P$ is $x^2 = 2y$. Let's check the given options: (A) $y^2 = 2x$ (B) $x^2 = 2y$ (C) $x^2 = y$ (D) $y^2 = x$

Our derived equation matches option (B).

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Tips and Common Mistakes to Avoid

1. Incorrect Vertex: Always double-check the vertex of the given parabola. For $x^2=4ay$ or $y^2=4ax$, it's $(0,0)$. If it were, for example, $(x-h)^2 = 4a(y-k)$, the vertex would be $(h,k)$.
2. Section Formula Ratio Confusion: Ensure you apply the ratio $m:n$ correctly. If $P$ divides $AB$ in ratio $m:n$, then $m$ multiplies the coordinates of $B$ and $n$ multiplies the coordinates of $A$.
3. Algebraic Errors: Be careful with squaring terms and simplifying equations. For instance, $(4h)^2 = 16h^2$, not $4h^2$.
4. Forgetting to Substitute: The most common mistake in locus problems is not fully eliminating the coordinates of the arbitrary point ($Q$ in this case) by using its defining equation.
5. Final Variable Change: Remember to change $(h,k)$ back to $(x,y)$ for the final equation of the locus.

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Summary and Key Takeaway

This problem demonstrates a standard procedure for finding the locus of a point:

1. Identify the fixed points/curves and the variable point whose locus is desired.
2. Assign general coordinates $(h,k)$ to the point whose locus is to be found.
3. Assign general coordinates $(x_1, y_1)$ (or parametric coordinates if easier) to any other variable point(s) involved in the definition of $(h,k)$.
4. Use the given geometric conditions (like section formula, distance formula, etc.) to establish equations relating $(h,k)$ and $(x_1, y_1)$.
5. Use the equation of the curve on which $(x_1, y_1)$ lies to eliminate $x_1$ and $y_1$, resulting in an equation purely in terms of $h$ and $k$.
6. Replace $(h,k)$ with $(x,y)$ to obtain the equation of the locus.

The new locus $x^2 = 2y$ is also a parabola, but it is "scaled down" or "compressed" compared to the original parabola $x^2 = 8y$. This makes sense because $P$ is always closer to the origin $O$ than $Q$ (since $P$ divides $OQ$ in ratio $1:3$, $OP = \frac{1}{4} OQ$).

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