Key Concepts and Formulas

To effectively solve this problem, we will leverage fundamental concepts from coordinate geometry, specifically concerning parabolas and lines. Understanding these foundational principles is crucial for a systematic approach.

1. Standard Equation of a Parabola: A parabola with its vertex at the origin and opening to the right has the standard equation $y^2 = 4ax$. Here, $a$ is a positive constant representing the focal length and determining the width of the parabola. This standard form is the basis for deriving tangent equations.

2. Equation of Tangent to a Parabola (Slope Form): For a parabola given by $y^2 = 4ax$, the equation of a tangent line with a specified slope $m$ (where $m \neq 0$) is given by: $y = mx + \frac{a}{m}$ This formula provides a direct and efficient way to find the tangent's equation if its slope and the parabola's parameter 'a' are known. It saves us from using more complex methods like finding the point of tangency first.

3. Slope of a Line: For a linear equation in the general form $Ax + By + C = 0$, its slope $M$ can be conveniently found by rearranging it into the slope-intercept form $y = Mx + C'$. In this form, $M = -A/B$. This conversion is a standard technique to quickly extract the slope from any linear equation.

4. Condition for Perpendicular Lines: If two lines are perpendicular to each other, the product of their slopes is $-1$. That is, if $m_1$ and $m_2$ are the slopes of two perpendicular lines, then $m_1 m_2 = -1$. This implies that one slope is the negative reciprocal of the other: $m_2 = -\frac{1}{m_1}$. This geometric property is fundamental for finding the slope of our tangent line, given the slope of a perpendicular line.

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Step-by-Step Solution

Let's break down the problem into manageable steps, explaining the rationale behind each action.

Step 1: Determine the Parameter 'a' for the Given Parabola

The first step is to identify the crucial parameter 'a' for the given parabola, as it is essential for using the standard tangent equation formula.

• Given Parabola Equation: $y^2 = 6x$
• Standard Parabola Equation (opening right): $y^2 = 4ax$

Why this step? The formula for the tangent to a parabola ($y = mx + a/m$) explicitly requires the value of $a$. By comparing the given equation with the standard form, we can accurately determine this parameter, which characterizes the specific parabola we are working with.

Comparing $y^2 = 6x$ with $y^2 = 4ax$: We equate the coefficients of $x$: $4a = 6$ Solving for $a$: $a = \frac{6}{4}$ $a = \frac{3}{2}$ So, for our parabola, the parameter $a$ is $\frac{3}{2}$.

Step 2: Find the Slope of the Line Perpendicular to the Tangent

The problem states that our desired tangent line is perpendicular to a given line. Therefore, we first need to determine the slope of this given line.

• Given Line Equation: $2x + y = 1$

Why this step? The condition for perpendicularity directly relates the slope of the given line to the slope of our tangent line. We need to know the former to calculate the latter. Without the slope of the reference line, we cannot proceed to find the tangent's slope.

To find the slope, we rearrange the equation into the slope-intercept form $y = Mx + C$: $y = -2x + 1$ From this form, we can clearly see that the slope of the given line, let's denote it as $m_{line}$, is: $m_{line} = -2$

Step 3: Calculate the Slope of the Tangent Line

Now that we have the slope of the line perpendicular to the tangent, we can use the perpendicularity condition to find the slope of the tangent itself.

Why this step? The slope of the tangent is a key component required to write its equation using the formula $y = mx + a/m$. The problem provides an indirect way to find this slope, and applying the perpendicularity condition is the direct path to it.

Let $m_T$ be the slope of the tangent line. Since the tangent is perpendicular to the line $2x + y = 1$, the product of their slopes must be $-1$: $m_T \cdot m_{line} = -1$ Substitute the value of $m_{line} = -2$ we found: $m_T \cdot (-2) = -1$ Solving for $m_T$: $m_T = \frac{-1}{-2} = \frac{1}{2}$ Thus, the slope of our desired tangent line is $\frac{1}{2}$.

Step 4: Formulate the Equation of the Tangent Line

With the parameter $a$ and the slope of the tangent $m_T$ now determined, we can substitute these values into the standard slope-form equation for a tangent to the parabola $y^2 = 4ax$.

Why this step? This is the central goal of the initial part of the problem: to find the specific equation of the tangent line that meets all the given criteria. Once we have this equation, we can test the given points to see which one does not lie on it.

The general equation of a tangent to $y^2 = 4ax$ with slope $m$ is: $y = mx + \frac{a}{m}$

Substitute $a = \frac{3}{2}$ (from Step 1) and $m = m_T = \frac{1}{2}$ (from Step 3): $y = \left(\frac{1}{2}\right)x + \frac{\frac{3}{2}}{\frac{1}{2}}$ Simplify the expression: $y = \frac{x}{2} + \left(\frac{3}{2} \cdot \frac{2}{1}\right)$ $y = \frac{x}{2} + 3$

To eliminate fractions and get a more standard linear equation form, which is often easier for checking points, we multiply the entire equation by 2: $2y = x + 6$ Rearrange the terms to get the equation in the general form $Ax + By + C = 0$: $x - 2y + 6 = 0$ This is the equation of the tangent line.

Step 5: Check Which Point Does NOT Lie on the Tangent

The final step is to test each of the given options by substituting their coordinates into the equation of the tangent line we just found. A point lies on the line if its coordinates satisfy the equation (i.e., make the equation true, resulting in $0=0$). We are looking for the point that does not satisfy it.

Why this step? This directly addresses the question asked. We systematically check each option to find the one that fails to satisfy the tangent line's equation, thereby identifying the point that does not lie on it.

Let's test each option with the tangent equation $x - 2y + 6 = 0$:

• (A) Point (0, 3): Substitute $x = 0$ and $y = 3$: $(0) - 2(3) + 6 = 0$ $0 - 6 + 6 = 0$ $0 = 0$ This is true. So, point (0, 3) does lie on the tangent.

• (B) Point (-6, 0): Substitute $x = -6$ and $y = 0$: $(-6) - 2(0) + 6 = 0$ $-6 - 0 + 6 = 0$ $0 = 0$ This is true. So, point (-6, 0) does lie on the tangent.

• (C) Point (4, 5): Substitute $x = 4$ and $y = 5$: $(4) - 2(5) + 6 = 0$ $4 - 10 + 6 = 0$ $-6 + 6 = 0$ $0 = 0$ This is true. So, point (4, 5) does lie on the tangent.

• (D) Point (5, 4): Substitute $x = 5$ and $y = 4$: $(5) - 2(4) + 6 = 0$ $5 - 8 + 6 = 0$ $-3 + 6 = 0$ $3 = 0$ This is false. So, point (5, 4) does NOT lie on the tangent.

Since the question asks for the point that does NOT lie on the tangent, the correct option is (D).

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Tips and Common Mistakes

• Correctly Identify 'a': A very common mistake is to confuse $4a$ with $a$. For $y^2 = 6x$, $a$ is not $6$; it's $6/4 = 3/2$. Always compare with the standard form $y^2 = 4ax$ carefully. Ensure you extract $a$ and not $4a$.
• Perpendicular vs. Parallel Slopes: Ensure you use the correct condition for slopes. For perpendicular lines, slopes are negative reciprocals ($m_1 m_2 = -1$). For parallel lines, slopes are equal ($m_1 = m_2$). Don't simply use the slope of the given line directly for the tangent.
• Algebraic Accuracy: Pay close attention to signs, fractions, and basic arithmetic throughout your calculations. A small error in determining 'a', the slope, or simplifying the equation can lead to a completely different tangent equation and an incorrect answer.
• Read the Question Carefully: The question asks for the point that does NOT lie on the tangent. It's easy to mistakenly select a point that does lie on it if you're not careful about the negation in the question.
• Check Your Work (Optional but Recommended): If time permits, you can quickly verify your tangent equation. For instance, substitute $y = \frac{x}{2} + 3$ back into the parabola equation $y^2 = 6x$. This gives $(\frac{x}{2} + 3)^2 = 6x$. Expanding this, you should get a quadratic equation in $x$ with a discriminant equal to zero. This confirms that the line is indeed tangent to the parabola at exactly one point.

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Summary and Key Takeaway

This problem is a classic application of analytical geometry principles for parabolas and lines. The solution path involves a sequence of logical steps:

1. Parameter Extraction: Accurately determine the parabola's key parameter 'a' by comparing its given equation with the standard form. This parameter is fundamental to the tangent equation.
2. Slope Determination: Find the slope of the reference line by converting its equation to slope-intercept form.
3. Condition Application: Use the geometric condition for perpendicular lines to deduce the slope of the desired tangent line from the slope of the reference line.
4. Equation Formation: Construct the tangent line's equation using the standard slope-form formula ($y = mx + a/m$) by substituting the calculated 'a' and 'm'.
5. Verification: Systematically test each of the given options by substituting their coordinates into the tangent's equation to identify the one that does not satisfy it, thus answering the question.

Mastering the standard forms of conic sections and the properties of lines (especially slope and conditions for perpendicularity/parallelism) is fundamental for success in JEE Mathematics. Practice ensures speed and accuracy in these types of problems, allowing you to quickly identify the correct formulas and apply them without error.