Here's a more elaborate, clear, and educational solution:

Key Concept: Shortest Distance from a Point to a Curve

When finding the point on a curve that is nearest to a given external point, a fundamental geometric principle applies: the normal to the curve at that nearest point must pass through the given external point.

Why this is true: Imagine a line segment connecting the external point $Q$ to a point $P$ on the curve. If $P$ is indeed the closest point, then this line segment $QP$ represents the shortest distance. Geometrically, the shortest distance from a point to a curve (or a line) is always along the perpendicular from the point to the curve (or line). Therefore, the line segment $QP$ must be perpendicular to the tangent line at point $P$. Since the normal to a curve at a point is defined as the line perpendicular to the tangent at that point, it follows that the line segment $QP$ must lie along the normal to the curve at $P$. In other words, the normal to the curve at the point of closest approach must pass through the external point $Q$. This property significantly simplifies the problem compared to directly minimizing the distance formula.

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1. Understanding the Parabola and its Parametric Representation

The given curve is a parabola with the equation $y^2 = 6x$. Our first step is to recognize its standard form and identify its key parameter.

• Standard Form: The general standard form of a parabola opening to the right is $y^2 = 4ax$.

• Identifying 'a': By comparing $y^2 = 6x$ with $y^2 = 4ax$, we can equate the coefficients of $x$: $4a = 6$ Solving for $a$: $a = \frac{6}{4} = \frac{3}{2}$ This value of $a$ is crucial for defining properties of this specific parabola.

• Parametric Representation: To simplify calculations for points on the parabola, especially when dealing with derivatives and tangents/normals, we use a parametric representation. For a parabola of the form $y^2 = 4ax$, any point $P$ on the curve can be conveniently represented by its parametric coordinates $(at^2, 2at)$, where $t$ is a real parameter. Why use parametric form? It allows us to represent both $x$ and $y$ coordinates using a single variable $t$, making differentiation and substitution much easier than working with $y = \pm\sqrt{6x}$.

• Specific Parametric Form for our Parabola: Now, we substitute the value $a = \frac{3}{2}$ into the general parametric form: $P \equiv \left( \frac{3}{2}t^2, 2 \cdot \frac{3}{2}t \right)$ $P \equiv \left( \frac{3}{2}t^2, 3t \right)$ The problem states that the nearest point is $(\alpha, \beta)$. Therefore, we can express $\alpha$ and $\beta$ in terms of $t$: $\alpha = \frac{3}{2}t^2 \quad \text{and} \quad \beta = 3t$ Our goal is now to find the value of $t$ that corresponds to the nearest point.

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2. Deriving the Equation of the Normal to the Parabola

To apply the key concept, we need the equation of the normal to the parabola at the point $P(\alpha, \beta)$, which we've represented parametrically as $P\left(\frac{3}{2}t^2, 3t\right)$.

Step 2.1: Find the slope of the tangent ($m_T$). We start by finding the derivative $\frac{dy}{dx}$, which gives the slope of the tangent at any point $(x,y)$ on the curve.

• Implicit Differentiation: Differentiate the equation of the parabola $y^2 = 6x$ implicitly with respect to $x$: $2y \frac{dy}{dx} = 6$ Why implicit? It's simpler than first solving for $y$ (e.g., $y = \sqrt{6x}$) and then differentiating.
• Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{6}{2y} = \frac{3}{y}$
• Evaluate at the parametric point: Now, substitute the $y$-coordinate of our parametric point $P(at^2, 2at)$, which is $2at$. Since $a=\frac{3}{2}$, the $y$-coordinate is $3t$. $m_T = \frac{3}{3t} = \frac{1}{t} \quad (\text{provided } t \neq 0)$
  • Special Case ($t=0$): If $t=0$, the point $P$ is $(0,0)$. At the origin, $y=0$, so $\frac{dy}{dx}$ is undefined, meaning the tangent is a vertical line ($x=0$, the y-axis). In this case, the normal would be a horizontal line ($y=0$, the x-axis). We will check if $t=0$ is relevant later.

Step 2.2: Find the slope of the normal ($m_N$). The normal line is perpendicular to the tangent line. The product of the slopes of two perpendicular lines is $-1$. $m_N = -\frac{1}{m_T}$ Substituting $m_T = \frac{1}{t}$: $m_N = -\frac{1}{1/t} = -t$

Step 2.3: Write the equation of the normal. We use the point-slope form of a line, $Y - y_1 = m_N (X - x_1)$, where $(x_1, y_1)$ is the point $P\left(\frac{3}{2}t^2, 3t\right)$ and $m_N = -t$.

• Substitute values: $Y - 3t = -t \left( X - \frac{3}{2}t^2 \right)$
• Simplify and rearrange: $Y - 3t = -tX + \frac{3}{2}t^3$ Rearranging to the standard form of the normal equation for $y^2=4ax$ (which is $tX + Y = 2at + at^3$): $tX + Y = 3t + \frac{3}{2}t^3$ This is the equation of the normal to the parabola $y^2 = 6x$ at any point $P\left(\frac{3}{2}t^2, 3t\right)$.

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3. Applying the Shortest Distance Condition

According to our key concept, for $P(\alpha, \beta)$ to be the nearest point on the parabola to the external point $Q\left(3, \frac{3}{2}\right)$, the normal to the parabola at $P$ must pass through $Q$. Therefore, we substitute the coordinates of the external point $(X, Y) = \left(3, \frac{3}{2}\right)$ into the equation of the normal we just derived: $t(3) + \frac{3}{2} = 3t + \frac{3}{2}t^3$

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4. Solving for the Parameter $t$ and Finding the Closest Point

Now, we solve this algebraic equation for $t$: $3t + \frac{3}{2} = 3t + \frac{3}{2}t^3$

• Simplify the equation: Notice that the term $3t$ appears on both sides of the equation. We can subtract $3t$ from both sides: $\frac{3}{2} = \frac{3}{2}t^3$
• Isolate $t^3$: To find $t^3$, divide both sides by $\frac{3}{2}$: $1 = t^3$
• Solve for $t$: Take the cube root of both sides. $t = \sqrt[3]{1}$ $t = 1$ Why only the real root? For geometric problems involving parameters like this, we are interested in real values of $t$ that correspond to actual points on the curve in the real coordinate plane. The complex roots of $t^3=1$ (i.e., $e^{i2\pi/3}$ and $e^{i4\pi/3}$) do not represent points on the real parabola.

Now that we have the value of $t=1$, we can find the coordinates of the nearest point $P(\alpha, \beta)$ by substituting $t=1$ into our parametric expressions from Section 1: $\alpha = \frac{3}{2}t^2 = \frac{3}{2}(1)^2 = \frac{3}{2}$ $\beta = 3t = 3(1) = 3$ So, the point on the curve $y^2 = 6x$ nearest to $\left(3, \frac{3}{2}\right)$ is $(\alpha, \beta) = \left(\frac{3}{2}, 3\right)$.

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5. Final Calculation

The question asks for the value of $2(\alpha + \beta)$. Substitute the values of $\alpha = \frac{3}{2}$ and $\beta = 3$: $2(\alpha + \beta) = 2\left(\frac{3}{2} + 3\right)$

• Add the terms inside the parenthesis: To add the fraction and the whole number, find a common denominator: $2\left(\frac{3}{2} + \frac{6}{2}\right)$ $= 2\left(\frac{3+6}{2}\right)$ $= 2\left(\frac{9}{2}\right)$
• Perform the multiplication: $= 9$

Thus, $2(\alpha + \beta) = 9$.

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Tips and Common Mistakes to Avoid

1. Misunderstanding the Normal Concept: This is the most critical part. Many students might try to use the distance formula directly, leading to a much more complex optimization problem involving square roots and derivatives, which is prone to algebraic errors. Always remember: the normal at the point of shortest distance passes through the external point.
2. Inefficient Parametric Form: Not using the parametric form $(at^2, 2at)$ for a parabola $y^2=4ax$ can make the differentiation and subsequent algebraic manipulations unnecessarily complicated. It's a powerful tool in coordinate geometry.
3. Incorrect Normal Equation: Ensure you correctly derive or recall the equation of the normal. For $y^2=4ax$, the normal at $(at^2, 2at)$ is $tX + Y = 2at + at^3$. If you forget it, practice deriving it quickly.
4. Algebraic Errors: Be very careful with arithmetic, especially when dealing with fractions. A small calculation mistake can propagate and lead to an incorrect final answer. Double-check each step.
5. Ignoring Special Cases: While not applicable here, always consider if $t=0$ or other specific parameter values might lead to undefined slopes or require separate analysis. In this problem, $t=1$ is a valid and straightforward solution.
6. Confusing Tangent and Normal: Remember that the normal's slope is the negative reciprocal of the tangent's slope.

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Summary and Key Takeaway

This problem beautifully demonstrates the power of combining geometric properties with algebraic techniques in coordinate geometry. The key insight is that the line connecting the external point to the nearest point on the curve must be perpendicular to the curve's tangent at that point (i.e., it must be the normal). By representing the parabola parametrically, deriving the equation of its normal, and then enforcing the condition that this normal passes through the given external point, we efficiently determined the parameter $t$. This allowed us to find the coordinates of the nearest point $(\alpha, \beta)$ and subsequently calculate the required expression $2(\alpha + \beta)$. Mastering parametric equations and the properties of tangents/normals is essential for solving such problems in JEE Mathematics.