This problem is a comprehensive test of your understanding of conic sections, specifically involving the properties of tangents to parabolas and normals to hyperbolas. We are tasked with a multi-step process: first, determine a specific point $(\alpha, \beta)$ on a given parabola using information about its tangent. Next, using these values, define a hyperbola and find the equation of its normal at another specified point. Finally, we must identify which of the given options does not lie on this calculated normal.

Let's break down the problem into manageable steps, applying the relevant formulas and concepts at each stage.

1. Step 1: Determining the Point $(\alpha, \beta)$ on the Parabola

Our first goal is to find the coordinates of the point $(\alpha, \beta)$ on the parabola $y^2 = 24x$. We are given that the tangent at this point is perpendicular to the line $2x+2y=5$.

• 1.1 Equation of Tangent to a Parabola

  • Key Concept: For a parabola of the form $y^2 = 4ax$, the equation of the tangent at a point $(x_1, y_1)$ on the parabola is given by $yy_1 = 2a(x+x_1)$. This is derived using differentiation or by the method of T-substitution.
  • Application: The given parabola is $y^2 = 24x$. Comparing this with the standard form $y^2 = 4ax$, we find $4a = 24$, which means $a = 6$.
  • The point of tangency is given as $(\alpha, \beta)$. Substituting $(x_1, y_1) = (\alpha, \beta)$ and $a=6$ into the tangent formula: $y\beta = 2(6)(x+\alpha)$ $y\beta = 12(x+\alpha)$ To find the slope of this tangent, we rearrange the equation into the slope-intercept form ($y=mx+c$): $y = \frac{12}{\beta}x + \frac{12\alpha}{\beta}$ So, the slope of the tangent at $(\alpha, \beta)$, denoted as $m_T$, is $m_T = \frac{12}{\beta}$.
  • Tip: It's crucial to correctly identify the parameter 'a' for the parabola and use the appropriate tangent formula. Misremembering the formula is a common mistake.

• 1.2 Slope of the Given Line

  • Key Concept: The slope of a linear equation $Ax+By=C$ is given by $m = -A/B$. Alternatively, rearrange the equation into $y=mx+c$ form.
  • Application: The given line is $2x+2y=5$. To find its slope, we rewrite it in slope-intercept form: $2y = -2x + 5$ $y = -x + \frac{5}{2}$ The slope of this line, denoted as $m_L$, is $m_L = -1$.

• 1.3 Using Perpendicularity Condition to Find Tangent's Slope

  • Key Concept: If two lines are perpendicular, the product of their slopes is -1 (i.e., $m_1 m_2 = -1$), provided neither line is vertical or horizontal.
  • Application: We are given that the tangent is perpendicular to the line $2x+2y=5$. Therefore, the product of their slopes must be -1: $m_T \cdot m_L = -1$ $\left(\frac{12}{\beta}\right) \cdot (-1) = -1$ $-\frac{12}{\beta} = -1$ Multiplying both sides by $-1$ and then by $\beta$, we get: $\frac{12}{\beta} = 1 \implies \beta = 12$

• 1.4 Finding the Coordinates $(\alpha, \beta)$

  • Key Concept: The point of tangency $(\alpha, \beta)$ must lie on the parabola itself, meaning its coordinates must satisfy the parabola's equation.
  • Application: We know $\beta = 12$ and the parabola's equation is $y^2 = 24x$. Substituting $(\alpha, \beta)$ into the equation: $\beta^2 = 24\alpha$ $(12)^2 = 24\alpha$ $144 = 24\alpha$ To find $\alpha$, divide both sides by 24: $\alpha = \frac{144}{24} = 6$
  • Thus, the point $(\alpha, \beta)$ is $(6, 12)$.
  • Common Mistake: Forgetting to use the parabola's equation to find the second coordinate after determining one of them. This step is essential to uniquely define the point.

2. Step 2: Defining the Hyperbola

Now that we have found $\alpha=6$ and $\beta=12$, we can write down the specific equation of the hyperbola.

• Key Concept: Substitute the determined values of $\alpha$ and $\beta$ into the given general equation of the hyperbola.
• Application: The hyperbola is given by $\frac{x^2}{\alpha^2} - \frac{y^2}{\beta^2} = 1$. Substituting $\alpha=6$ and $\beta=12$: $$$$