Key Concept: Equation of a Normal to a Parabola in Slope Form

For a standard parabola $y^2 = 4Ax$, where $A$ is the semi-latus rectum parameter (focal length), the equation of a normal with slope $m$ is given by: $y = mx - 2Am - Am^3$ This formula is derived from the parametric form of the parabola $(At^2, 2At)$. The slope of the tangent at $(At^2, 2At)$ is $1/t$. Therefore, the slope of the normal is $-t$. If we let $m = -t$, then $t = -m$. Substituting $t=-m$ into the point-slope form of a line $Y - 2At = (-t)(X - At^2)$ gives the general normal equation in terms of $m$.

For a parabola shifted horizontally, like $Y^2 = 4B(X-c)$, the formula adapts by replacing $y$ with $Y$, $x$ with $X-c$, and $A$ with $B$. The general form remains consistent.

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Problem Analysis: Finding a Common Normal

The problem asks us to find a valid choice for the ordered triad $(a, b, c)$ such that the two given parabolas, $y^2 = 4b(x – c)$ and $y^2 = 8ax$, have a common normal. A common normal implies that there exists a single straight line that is perpendicular to the tangent of both parabolas at their respective points of intersection. If such a line exists, it must have the same slope $m$ and the same $y$-intercept for both parabolas.

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Step-by-Step Solution

Step 1: Identify the Parameters for Each Parabola

Before applying the normal formula, we must correctly identify the effective 'A' parameter for each parabola. This parameter is crucial as it dictates the shape and orientation of the parabola, and thus the properties of its normals.

• Parabola 1: $y^2 = 4b(x - c)$ This parabola is a horizontal shift of $y^2 = 4bx$. Comparing it to the general form $Y^2 = 4AX$, we can see:

  • The variable corresponding to $Y$ is $y$.
  • The variable corresponding to $X$ is $(x-c)$.
  • The focal length parameter, which is 'A' in our normal formula, is $b$. So, for this parabola, the parameter to be used in the normal formula is $A_1 = b$.

• Parabola 2: $y^2 = 8ax$ This is a standard parabola with its vertex at the origin. Comparing it to the general form $y^2 = 4AX$, we have:

  • $4A = 8a$
  • Therefore, the focal length parameter, which is 'A' in our normal formula, is $A_2 = 2a$. So, for this parabola, the parameter to be used in the normal formula is $A_2 = 2a$.

  Why this step? Correctly identifying the 'A' parameter is fundamental. A common mistake is to use $4b$ or $8a$ directly in the formula, which would lead to incorrect results. The formula uses the semi-latus rectum parameter, not the coefficient of $x$.

Step 2: Write the Equation of the Normal for Each Parabola

Now, we apply the general normal formula $y = mx - 2Am - Am^3$ to each parabola, using the specific parameters identified in Step 1. We assume that a common normal exists and has a slope $m$.

• Normal to Parabola 1 ($y^2 = 4b(x - c)$): We use the adapted formula where $y$ is $y$, $x$ is $(x-c)$, and $A$ is $b$: $y = m(x - c) - 2bm - bm^3 \quad \text{(Equation 1)}$ This equation describes any normal line with slope $m$ to the first parabola.

• Normal to Parabola 2 ($y^2 = 8ax$): We use the standard formula where $y$ is $y$, $x$ is $x$, and $A$ is $2a$: $y = mx - 2(2a)m - (2a)m^3$ $y = mx - 4am - 2am^3 \quad \text{(Equation 2)}$ This equation describes any normal line with slope $m$ to the second parabola.

  Why this step? By writing these two equations, we are expressing the conditions for a line to be normal to each parabola individually. The next step will combine these conditions to find when they represent the same line.

Step 3: Equate the Normal Equations to Find the Condition for a Common Normal

For a line to be a common normal to both parabolas, the equations representing the normal to each parabola (Equation 1 and Equation 2) must be identical. Since we've already assumed they share the same slope $m$, their y-intercepts (or the constant terms when rearranged into $y=mx+C$ form) must also be identical.

Equating the expressions for $y$ from Equation 1 and Equation 2: $m(x - c) - 2bm - bm^3 = mx - 4am - 2am^3$

First, expand the left side: $mx - mc - 2bm - bm^3 = mx - 4am - 2am^3$

Now, subtract $mx$ from both sides, as the $mx$ term is common to both equations, meaning their slopes are already matched: $-mc - 2bm - bm^3 = -4am - 2am^3$

To make the equation easier to analyze, move all terms to one side and group them by powers of $m$: $4am + 2am^3 - mc - 2bm - bm^3 = 0$ $(4a - c - 2b)m + (2a - b)m^3 = 0$

Factor out $m$ from the entire expression: $m[(4a - c - 2b) + (2a - b)m^2] = 0$

Step 4: Analyze the Slope $m$ and Derive the Condition for a Real Common Normal

From the factored equation $m[(4a - c - 2b) + (2a - b)m^2] = 0$, we have two possibilities for the slope $m$:

• Case 1: $m = 0$ If $m=0$, the equation of the normal becomes $y = 0$. This is the x-axis. The x-axis is indeed a normal to $y^2 = 4b(x-c)$ at its vertex $(c,0)$ and to $y^2 = 8ax$ at its vertex $(0,0)$. Thus, $y=0$ is always a common normal to these types of parabolas, provided the vertices are distinct or coincide appropriately. In multiple-choice questions, when asked for a "common normal," it usually implies a non-trivial one (i.e., $m \neq 0$), unless all options lead to $m=0$ or no other real $m$.

• Case 2: $m \neq 0$ If $m \neq 0$, we can divide the equation by $m$: $(4a - c - 2b) + (2a - b)m^2 = 0$ Rearrange this equation to solve for $m^2$: $(2a - b)m^2 = -(4a - c - 2b)$ $m^2 = \frac{-(4a - c - 2b)}{2a - b}$ $m^2 = \frac{c + 2b - 4a}{2a - b}$

  For a real common normal with a non-zero slope to exist, $m$ must be a real number. This implies that $m^2$ must be strictly positive ($m^2 > 0$). If $m^2=0$, then $m=0$, which falls under Case 1. If $m^2$ is negative, there is no real slope $m$, and thus no real common normal (other than possibly the x-axis). Therefore, the condition for a non-trivial (i.e., $m \neq 0$) real common normal is: $\frac{c + 2b - 4a}{2a - b} > 0$ Also, we must ensure that the denominator $2a-b \neq 0$. If $2a-b=0$, the equation becomes $4a-c-2b=0$, which means $(2a-b)m^2 = 0$ and $m$ can be anything as long as $4a-c-2b=0$. This implies infinite normals or no common normal, depending on the specifics. For a unique $m^2$, $2a-b$ must be non-zero.

  Why this step? This step is critical for isolating the condition on the parameters $a, b, c$. By analyzing the equation for $m$, we determine when a real and non-trivial common normal can exist. This is usually what JEE questions test.

Step 5: Test the Given Options

Now, we substitute the values of $(a, b, c)$ from each option into the derived condition for $m^2$: $m^2 = \frac{c + 2b - 4a}{2a - b}$ We are looking for the option where $m^2 > 0$.

• (A) $(1, 1, 3)$ Substitute $a=1, b=1, c=3$: $m^2 = \frac{3 + 2(1) - 4(1)}{2(1) - 1} = \frac{3 + 2 - 4}{2 - 1} = \frac{1}{1} = 1$ Since $m^2 = 1 > 0$, this is a valid choice. This means there are two real common normals with slopes $m = \pm 1$.

• (B) $(1, 1, 0)$ Substitute $a=1, b=1, c=0$: $m^2 = \frac{0 + 2(1) - 4(1)}{2(1) - 1} = \frac{2 - 4}{1} = -2$ Since $m^2 = -2$, which is not greater than $0$, this option does not yield a real common normal with $m \neq 0$.

• (C) $\left( \frac{1}{2}, 2, 0 \right)$ Substitute $a=\frac{1}{2}, b=2, c=0$: $m^2 = \frac{0 + 2(2) - 4\left(\frac{1}{2}\right)}{2\left(\frac{1}{2}\right) - 2} = \frac{4 - 2}{1 - 2} = \frac{2}{-1} = -2$ Since $m^2 = -2$, which is not greater than $0$, this option does not yield a real common normal with $m \neq 0$.

• (D) $\left( \frac{1}{2}, 2, 3 \right)$ Substitute $a=\frac{1}{2}, b=2, c=3$: $m^2 = \frac{3 + 2(2) - 4\left(\frac{1}{2}\right)}{2\left(\frac{1}{2}\right) - 2} = \frac{3 + 4 - 2}{1 - 2} = \frac{5}{-1} = -5$ Since $m^2 = -5$, which is not greater than $0$, this option does not yield a real common normal with $m \neq 0$.

  Why this step? This is the final verification. By systematically checking each option against the derived condition, we definitively identify the correct answer choice. Option (A) is the only one that satisfies the condition for a real, non-trivial common normal.

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Tips for Success and Common Pitfalls

• Parameter Identification is Key: The most common mistake is incorrectly identifying the 'A' parameter for the normal formula $y = mx - 2Am - Am^3$. Always ensure you convert $y^2 = KX$ to $y^2 = 4AX$, so $A = K/4$.
  • For $y^2 = 4b(x-c)$, the parameter is $b$.
  • For $y^2 = 8ax$, the parameter is $2a$.
• Common Normal Logic: For two lines to be identical, both their slopes and their constant terms (y-intercepts) must match. In this problem, we assumed a common slope $m$ and then equated the constant terms.
• The $m=0$ Case: While $