Understanding the Foundation: Parabola and its Normal

This problem requires a solid understanding of the parametric representation of a parabola and the equation of its normal line. These are fundamental concepts in coordinate geometry, frequently tested in JEE.

The standard equation of a parabola is $y^2 = 4ax$. A point on this parabola can be conveniently represented in parametric form as $P(at^2, 2at)$, where $t$ is the parameter. This form simplifies many calculations involving tangents and normals.

For this specific problem, the parabola is given by points of the form $(bt^2, 2bt)$. This implies our parabola's equation is $y^2 = 4bx$, where the constant 'a' from the standard form is replaced by 'b'.

We are given two points on this parabola:

1. Point $P(bt_1^2, 2bt_1)$, associated with parameter $t_1$.
2. Point $Q(bt_2^2, 2bt_2)$, associated with parameter $t_2$.

The critical tool for this problem is the equation of the normal to a parabola at a given parametric point. For a parabola $y^2 = 4ax$, the equation of the normal at the point $(at^2, 2at)$ is: $y = -tx + 2at + at^3$

Step-by-Step Derivation of the Relationship between $t_1$ and $t_2$

Our goal is to find a relationship between $t_1$ and $t_2$ based on the condition that the normal at $P$ passes through $Q$.

Step 1: Write down the equation of the normal at point $P(bt_1^2, 2bt_1)$.

• Why this step? The problem states that "the normal at the point $P$..." This is our starting line. We need the algebraic representation of this normal line.
• Action: We adapt the general normal equation $y = -tx + 2at + at^3$ by replacing 'a' with 'b' and 't' with $t_1$ (since the normal is drawn at $P$ with parameter $t_1$). $y = -t_1x + 2bt_1 + bt_1^3 \quad \text{... (1)}$ This is the equation of the straight line perpendicular to the parabola at point $P$.

Step 2: Apply the condition that the normal meets the parabola again at point $Q(bt_2^2, 2bt_2)$.

• Why this step? The problem specifies that point $Q$ lies on the normal line we just found. If a point lies on a line, its coordinates must satisfy the equation of that line. By substituting the coordinates of $Q$ into equation (1), we establish a direct algebraic link between $t_1$ and $t_2$.
• Action: Substitute $x = bt_2^2$ and $y = 2bt_2$ into equation (1): $2bt_2 = -t_1(bt_2^2) + 2bt_1 + bt_1^3$

Step 3: Simplify the equation by dividing by 'b'.

• Why this step? To make the equation easier to manipulate algebraically, we look for common factors. Every term in the equation contains 'b'. Since 'b' is a parameter defining the parabola (and for a non-degenerate parabola, $b \neq 0$), we can safely divide the entire equation by 'b' without losing any solutions or invalidating the equation.
• Action: Divide every term by 'b': $2t_2 = -t_1t_2^2 + 2t_1 + t_1^3$

Step 4: Rearrange terms to facilitate factorization.

• Why this step? Our ultimate goal is to isolate $t_2$ and express it in terms of $t_1$. Algebraic factorization is a powerful technique to simplify expressions and reveal relationships. By moving terms around, we can group them to find common factors.
• Action: Move all terms involving $t_1$ from the right-hand side to the left-hand side, keeping similar terms together for easier factoring: $2t_2 - 2t_1 = -t_1t_2^2 + t_1^3$ Now, factor out common terms from both sides: $2(t_2 - t_1) = -t_1(t_2^2 - t_1^2)$

Step 5: Apply the difference of squares identity.

• Why this step? The term $(t_2^2 - t_1^2)$ is a standard algebraic identity: $A^2 - B^2 = (A-B)(A+B)$. Recognizing and applying this identity is crucial because it will introduce the factor $(t_2 - t_1)$ on the right side, which matches the factor already present on the left side. This is a common strategy in such problems.
• Action: Substitute $(t_2 - t_1)(t_2 + t_1)$ for $(t_2^2 - t_1^2)$: $2(t_2 - t_1) = -t_1(t_2 - t_1)(t_2 + t_1)$

Step 6: Divide by $(t_2 - t_1)$.

• Why this step? We have a common factor $(t_2 - t_1)$ on both sides. To simplify the equation further, we can divide by this factor.
• Important Consideration: We must ensure that the term we are dividing by is not zero. The problem states that the normal meets the parabola again in the point $Q$. This explicitly implies that $Q$ is a distinct point from $P$. If $t_2 = t_1$, then $P$ and $Q$ would be the same point, which contradicts the problem statement of meeting "again". Therefore, $t_2 \neq t_1$, which means $(t_2 - t_1) \neq 0$. This allows us to safely divide by $(t_2 - t_1)$ without losing a valid solution.
• Action: Divide both sides by $(t_2 - t_1)$: $2 = -t_1(t_2 + t_1)$

Step 7: Isolate $t_2$ to find the final relationship.

• Why this step? We have successfully simplified the equation to a point where we can directly solve for $t_2$ in terms of $t_1$, which is the desired relationship.
• Action: First, divide both sides by $-t_1$: $\frac{2}{-t_1} = t_2 + t_1$ $-\frac{2}{t_1} = t_2 + t_1$ Finally, subtract $t_1$ from both sides to isolate $t_2$: $t_2 = -t_1 - \frac{2}{t_1}$

This derived relationship matches option (B).

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JEE Aspirant Tips and Common Mistakes to Avoid

1. Memorize Key Formulas: For conic sections, especially parabolas, having the parametric forms, equations of tangents, and equations of normals readily memorized is crucial. Deriving them during the exam is time-consuming.
   • Parabola $y^2 = 4ax$: Point $(at^2, 2at)$, Tangent $ty = x + at^2$, Normal $y = -tx + 2at + at^3$.
2. Careful Algebraic Manipulations:
   • Signs: Pay close attention to negative signs, especially when rearranging terms or factoring. A single sign error can lead to the wrong option.
   • Factoring: Be adept at recognizing and applying algebraic identities like the difference of squares.
   • Dividing by Variables: Always consider the condition under which you can divide by a variable term. Here, $t_2 - t_1 \neq 0$ because the points are distinct. If the question implied the normal was a tangent (meeting at the same point), then $t_1=t_2$ would be a valid case, and you couldn't simply divide.
3. Understand Geometric Context: The phrase "meets the parabola again" is a strong hint. It confirms that the two points $P$ and $Q$ are distinct, which algebraically means $t_1 \neq t_2$.
4. Recognize Standard Results: The relationship $t_2 = -t_1 - \frac{2}{t_1}$ is a standard and frequently encountered result for a normal to a parabola $y^2=4ax$ meeting the parabola again. While knowing the derivation is essential, recognizing this result can boost confidence and save time if you're ever in a hurry.

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Summary and Key Takeaway

This problem is a classic application of the parametric form of a parabola and its normal equation. The key steps involve:

1. Writing the equation of the normal at the first point $P(bt_1^2, 2bt_1)$.
2. Substituting the coordinates of the second point $Q(bt_2^2, 2bt_2)$ into the normal's equation.
3. Carefully simplifying the resulting equation using algebraic techniques, particularly factorization and the difference of squares identity.
4. Ensuring that division by terms like $(t_2 - t_1)$ is valid due to the distinct nature of points $P$ and $Q$.

The final relationship derived is $t_2 = -t_1 - \frac{2}{t_1}$. This is a standard and important result for JEE preparation.

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