Problem Statement: If $y = mx + c$ is the normal at a point on the parabola $y^2 = 8x$ whose focal distance is 8 units, then $\left| c \right|$ is equal to:

Understanding the Problem: We are given a parabola and a specific point on it, defined by its focal distance. At this point, a normal line is drawn, and its equation is given in the slope-intercept form $y = mx + c$. Our goal is to find the absolute value of the y-intercept, $|c|$. This problem requires us to connect the parabola's equation, its focal properties, and the equation of its normal.

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Key Concepts and Formulas for Parabola $y^2 = 4ax$

To solve this problem effectively, we will utilize the following fundamental concepts and formulas related to parabolas:

1. Standard Equation of a Parabola: The equation of a parabola opening to the right with its vertex at the origin is given by $y^2 = 4ax$. Here, '$a$' is a crucial parameter. It represents:

   • The distance from the vertex to the focus $F(a, 0)$.
   • The distance from the vertex to the directrix $x = -a$.

2. Parametric Coordinates of a Point: Any point $P$ on the parabola $y^2 = 4ax$ can be conveniently represented using a parameter 't' as $P(at^2, 2at)$. This form simplifies calculations involving properties of points on the parabola.

3. Focal Distance: For any point $P(x_1, y_1)$ on the parabola $y^2 = 4ax$, its focal distance (the distance from the point to the focus $F(a, 0)$) is given by $x_1 + a$. In parametric form, for $P(at^2, 2at)$, the focal distance is $at^2 + a = a(t^2 + 1)$. This property arises directly from the definition of a parabola: every point on the parabola is equidistant from its focus and its directrix.

4. Equation of the Normal: The equation of the normal to the parabola $y^2 = 4ax$ at the point $P(at^2, 2at)$ in parametric form is: $y + tx = 2at + at^3$ This equation is derived by first finding the slope of the tangent at $P$ (which is $1/t$) and then using the negative reciprocal for the slope of the normal (which is $-t$).

5. Y-intercept of a Line: For a linear equation expressed in the slope-intercept form $y = mx + c$, the constant 'c' represents the y-intercept. This is the y-coordinate of the point where the line crosses the y-axis (i.e., when $x=0$).

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Step-by-Step Solution

Step 1: Determine the Parabola's Parameter 'a'

• Why this step is important: The parameter 'a' is fundamental to defining the specific parabola we are working with. All subsequent calculations, including the focus, directrix, focal distance, and the equations of tangent/normal, depend on this value. Therefore, identifying 'a' is always the first crucial step when dealing with a specific parabola equation.

• Working: We are given the equation of the parabola as $y^2 = 8x$. We compare this with the standard form of a parabola opening to the right, which is $y^2 = 4ax$. By equating the coefficients of $x$ from both equations: $4a = 8$ Now, we solve for 'a': $a = \frac{8}{4}$ $a = 2$ Thus, for the given parabola $y^2 = 8x$, the parameter $a$ is 2.

Step 2: Use Focal Distance to Find the Point Parameter 't'

• Why this step is important: The problem specifies that the normal is drawn at a point whose focal distance is 8 units. This information is key to identifying the exact point on the parabola. In the parametric representation, this point is uniquely determined by its parameter 't'. Finding 't' allows us to then write the equation of the normal at that specific point.

• Working: The focal distance of a point $P(at^2, 2at)$ on the parabola $y^2 = 4ax$ is given by the formula $a(t^2 + 1)$. We are given that the focal distance is 8 units. From Step 1, we found $a=2$. Substitute these values into the focal distance formula: $a(t^2 + 1) = 8$ $2(t^2 + 1) = 8$ Now, we solve for $t^2$: $t^2 + 1 = \frac{8}{2}$ $t^2 + 1 = 4$ $t^2 = 4 - 1$ $t^2 = 3$ Taking the square root, we find the possible values for $t$: $t = \pm \sqrt{3}$ For the purpose of finding $|c|$, choosing $t = \sqrt{3}$ or $t = -\sqrt{3}$ will result in $c$ values that are negatives of each other (e.g., $10\sqrt{3}$ and $-10\sqrt{3}$), but their absolute values will be the same. Let's proceed with $t = \sqrt{3}$ for our calculation.

Step 3: Determine the Y-intercept 'c' of the Normal

• Why this step is important: The question asks for $|c|$, where $y = mx + c$ is the equation of the normal. We now have all the necessary information ($a$ and $t$) to write the specific equation of the normal at the identified point. By expressing this normal equation in the slope-intercept form, we can directly read off the value of 'c'.

• Working: The equation of the normal to the parabola $y^2 = 4ax$ at the point $(at^2, 2at)$ in parametric form is: $y + tx = 2at + at^3$ To find the y-intercept 'c', we need to express this equation in the slope-intercept form $y = mx + c$. We do this by isolating $y$: $y = -tx + (2at + at^3)$ Comparing this with $y = mx + c$, we can clearly identify the y-intercept $c$ as the constant term: $c = 2at + at^3$ We can factor out $at$ from the expression for $c$ to simplify calculations: $c = at(2 + t^2)$ Now, substitute the values we found: $a=2$ (from Step 1) and $t=\sqrt{3}$ (from Step 2): $c = (2)(\sqrt{3})(2 + (\sqrt{3})^2)$ $c = 2\sqrt{3}(2 + 3)$ $c = 2\sqrt{3}(5)$ $c = 10\sqrt{3}$ (If we had chosen $t = -\sqrt{3}$, the calculation would be $c = (2)(-\sqrt{3})(2 + (-\sqrt{3})^2) = -2\sqrt{3}(2+3) = -10\sqrt{3}$. As expected, the absolute value will be the same.)

Step 4: Calculate $|c|$

• Why this step is important: The problem explicitly asks for the absolute value of $c$, not $c$ itself. This is the final step to provide the requested answer.

• Working: We found $c = 10\sqrt{3}$. The absolute value of $c$ is: $|c| = |10\sqrt{3}|$ Since $10\sqrt{3}$ is a positive value: $|c| = 10\sqrt{3}$

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Tips for JEE Aspirants & Common Mistakes to Avoid

• Parameter 'a' Identification: Always ensure you correctly extract the value of 'a' from the given parabola equation. For $y^2 = 8x$, $4a=8 \implies a=2$. A common mistake is to directly use 8 as 'a' or confuse $y^2=4ax$ with $x^2=4ay$.
• Focal Distance Formula: Remember that the focal distance for a point $(x_1, y_1)$ on $y^2=4ax$ is $x_1+a$. Don't just use $x_1$. The parametric form $a(t^2+1)$ is particularly useful.
• Tangent vs. Normal: Be extremely careful to distinguish between the equation of the tangent and the equation of the normal. They are different lines with slopes that are negative reciprocals of each other.
  • Equation of Tangent at $P(at^2, 2at)$: $ty = x + at^2$
  • Equation of Normal at $P(at^2, 2at)$: $y + tx = 2at + at^3$
• Parametric Form Efficiency: The parametric form $(at^2, 2at)$ is often the most efficient way to solve problems involving points on a parabola, especially when dealing with properties like focal distance and the equations of lines related to the parabola.
• Sign of 't' for Absolute Value: When a problem asks for an absolute value (like $|c|$), and an intermediate step yields $\pm$ values (like $t = \pm \sqrt{3}$), understand that both choices should lead to the same final absolute value. This can save time by allowing you to pick the positive value for 't' for simpler calculations.
• Algebraic Precision: Double-check your algebraic manipulations, especially when substituting values and simplifying expressions. A small calculation error can lead to a wrong answer.

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Summary and Key Takeaway

This problem is a quintessential example of how to systematically approach questions involving geometric properties of parabolas. The key steps are:

1. Standardization: Always begin by comparing the given parabola equation with its standard form to determine the crucial parameter 'a'.
2. Point Identification: Use the given information (in this case, focal distance) to precisely locate the point on the parabola where the geometric property (normal) is being considered. This usually involves finding the parameter 't'.
3. Equation Formulation: With 'a' and 't' identified, write the equation of the required line (the normal in this problem).
4. Final Calculation: Extract the requested information (the y-intercept 'c' and its absolute value) from the line's equation.

By following these structured steps, you can confidently solve complex problems by breaking them down into manageable and logical parts.

The final answer is \boxed{\text{10\sqrt 3}}.