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Step 1: Express the Point on the Parabola in Parametric Form
The given parabola is y^2 = 6x . A convenient parametrization for this curve is:
x = \tfrac{3}{2} t^2, \quad y = 3t,
for a real parameter t . Any point P on the parabola can thus be written as
P \bigl(\tfrac{3}{2} t^2, \, 3t\bigr) .
Step 2: Write the Equation of the Normal at the Point P
Using properties of the parabola y^2 = 6x , one can show that the equation of the normal at
P \bigl(\tfrac{3}{2} t^2, 3t\bigr) can be expressed as:
t \, x \;+\; y \;=\; 3t \;+\; \tfrac{3}{2} t^3.
Step 3: Impose the Condition that the Normal Passes through (3,\tfrac{3}{2})
We want the normal line at P to pass through the external point (3, \tfrac{3}{2}) . Substituting
x = 3 and y = \tfrac{3}{2} into the normal’s equation gives:
t \cdot 3 + \tfrac{3}{2} \;=\; 3t \;+\; \tfrac{3}{2} t^3.
This simplifies to:
3t + \tfrac{3}{2} = 3t + \tfrac{3}{2} t^3
\;\;\Longrightarrow\;\; \tfrac{3}{2} = \tfrac{3}{2} t^3
\;\;\Longrightarrow\;\; t^3 = 1
\;\;\Longrightarrow\;\; t = 1.
Step 4: Compute the Coordinates of the Required Point
With t = 1 , the coordinates of the point P on the parabola become:
\alpha = \tfrac{3}{2} (1)^2 = \tfrac{3}{2}, \quad \beta = 3 \times 1 = 3.
Hence, P(\alpha,\beta) = \bigl(\tfrac{3}{2}, 3\bigr).
Step 5: Find 2(\alpha + \beta)
We have:
\alpha + \beta = \tfrac{3}{2} + 3 = \tfrac{3}{2} + \tfrac{6}{2} = \tfrac{9}{2}.
Therefore,
2(\alpha + \beta)
= 2 \left(\tfrac{9}{2}\right)
= 9.
Answer
2(\alpha + \beta) = 9.
