This solution will guide you through finding the ordinate of point Q by systematically applying the properties of parabolas, specifically focusing on the standard form $y^2 = 4ax$.

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Understanding the Parabola $y^2 = 6x$ and its Properties

Before we begin, let's establish the fundamental equations and concepts for a parabola of the form $y^2 = 4ax$. These are crucial for solving problems involving tangents and normals.

1. Standard Form: The given parabola is $y^2 = 6x$. We compare this to the standard form $y^2 = 4ax$ to identify the parameter 'a'.
2. Parametric Point P: A point P on the parabola can be conveniently represented as $P(at^2, 2at)$, where 't' is a real parameter. This form simplifies calculations for tangents and normals.
3. Equation of Tangent at P: The equation of the tangent line to the parabola at $P(at^2, 2at)$ is $yt = x + at^2$.
4. Equation of Normal at P: The equation of the normal line (perpendicular to the tangent) to the parabola at $P(at^2, 2at)$ is $y = -tx + 2at + at^3$.
5. Equation of Directrix: The directrix of the parabola $y^2 = 4ax$ is the vertical line $x = -a$.

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Step-by-Step Solution

We will now apply these concepts to solve the problem systematically.

Step 1: Determine the Parabola's Fundamental Parameter 'a' and Parametric Point P

• Concept: The first step in any parabola problem is to identify its specific parameter 'a' by comparing it with the standard form. This 'a' value defines all geometric properties of the parabola.
• Given Parabola: We are given the equation $y^2 = 6x$.
• Comparison with Standard Form ($y^2 = 4ax$): By comparing the coefficients of $x$, we have: $4a = 6$ Solving for $a$: $a = \frac{6}{4} = \frac{3}{2}$
• Parametric Point P: Now that we have the value of $a$, we can express any point P on this parabola in its parametric form $P(at^2, 2at)$. This form is chosen because it simplifies the derivation of tangent and normal equations, making further calculations much easier. Substitute $a = \frac{3}{2}$ into the parametric coordinates: $P\left(\frac{3}{2}t^2, 2 \cdot \frac{3}{2}t\right)$ This simplifies to: $P\left(\frac{3}{2}t^2, 3t\right)$ This point P is where the normal is drawn.

Step 2: Formulate the Equation of the Normal at Point P

• Concept: The problem states that the normal at P passes through a specific point $(5, -8)$. To use this information, we first need the algebraic equation of the normal line in terms of 't' and 'a'.
• Standard Normal Equation: The equation of the normal to the parabola $y^2 = 4ax$ at $P(at^2, 2at)$ is: $y = -tx + 2at + at^3$
• Substitute 'a': Now, substitute the specific value of $a = \frac{3}{2}$ into this equation: $y = -tx + 2\left(\frac{3}{2}\right)t + \left(\frac{3}{2}\right)t^3$ Simplifying, we get the equation of the normal for our parabola: $y = -tx + 3t + \frac{3}{2}t^3 \quad (*)$

Step 3: Utilize the Given Point to Find the Parameter 't'

• Concept: The problem states that the normal passes through the point $(5, -8)$. If a point lies on a line, its coordinates must satisfy the line's equation. By substituting these coordinates into equation $(*)$, we can form an equation solely in terms of 't' and solve for it. Finding 't' will uniquely determine the point P.
• Substitute Point into Normal Equation: Substitute $x=5$ and $y=-8$ into equation $(*)$: $-8 = -t(5) + 3t + \frac{3}{2}t^3$ $-8 = -5t + 3t + \frac{3}{2}t^3$ $-8 = -2t + \frac{3}{2}t^3$
• Rearrange into a Cubic Equation: To work with integer coefficients, multiply the entire equation by 2: $-16 = -4t + 3t^3$ Rearranging the terms into a standard cubic polynomial form ($At^3 + Bt^2 + Ct + D = 0$): $3t^3 - 4t + 16 = 0$
• Solving the Cubic Equation (Finding 't'): We need to find the real roots of this cubic equation.
  • Tip: Rational Root Theorem: For cubic equations with integer coefficients, we can test integer factors of the constant term (16) divided by factors of the leading coefficient (3) to find rational roots. Let's test integer factors of 16: $\pm 1, \pm 2, \pm 4, \dots$
  • Test $t=1$: $3(1)^3 - 4(1) + 16 = 3 - 4 + 16 = 15 \neq 0$.
  • Test $t=-1$: $3(-1)^3 - 4(-1) + 16 = -3 + 4 + 16 = 17 \neq 0$.
  • Test $t=2$: $3(2)^3 - 4(2) + 16 = 3(8) - 8 + 16 = 24 - 8 + 16 = 32 \neq 0$.
  • Test $t=-2$: $3(-2)^3 - 4(-2) + 16 = 3(-8) + 8 + 16 = -24 + 8 + 16 = 0$. Since $t=-2$ satisfies the equation, $(t+2)$ is a factor of the cubic polynomial.
• Factor the Cubic: We can perform polynomial division or synthetic division to factor out $(t+2)$: $(t+2)(3t^2 - 6t + 8) = 0$
• Check for Other Real Roots: Now, we examine the quadratic factor $3t^2 - 6t + 8 = 0$. We need to determine if it has any other real roots. We do this by calculating its discriminant, $\Delta = b^2 - 4ac$: $\Delta = (-6)^2 - 4(3)(8)$ $\Delta = 36 - 96$ $\Delta = -60$ Since the discriminant is negative ($\Delta < 0$), the quadratic equation $3t^2 - 6t + 8 = 0$ has no real roots (it has two complex conjugate roots).
• Conclusion for 't': Therefore, the only real value for the parameter $t$ is $t = -2$. This uniquely identifies the point P on the parabola where the normal is drawn.

Step 4: Determine the Equation of the Tangent at P and the Directrix

• Concept: The problem asks for the ordinate of point Q, which is the intersection of the tangent at P and the directrix. Thus, we need the equations of both these lines.
• Equation of Tangent at P: The equation of the tangent to $y^2 = 4ax$ at $P(at^2, 2at)$ is $yt = x + at^2$. Substitute $a = \frac{3}{2}$ and the value of $t = -2$ that we just found: $y(-2) = x + \frac{3}{2}(-2)^2$ $-2y = x + \frac{3}{2}(4)$ $-2y = x + 6 \quad (**)$
• Equation of Directrix: The directrix for the parabola $y^2 = 4ax$ is given by $x = -a$. Substitute $a = \frac{3}{2}$: $x = -\frac{3}{2}$

Step 5: Find the Ordinate of the Intersection Point Q

• Concept: Point Q is the intersection of the tangent $(**)$ and the directrix. To find the coordinates of Q, we substitute the x-coordinate of the directrix into the tangent equation. We are specifically looking for the ordinate (y-coordinate) of Q.

• Substitute Directrix into Tangent Equation: Substitute $x = -\frac{3}{2}$ into the tangent equation $(**)$: $-2y_Q = \left(-\frac{3}{2}\right) + 6$ To simplify the right side, find a common denominator: $-2y_Q = -\frac{3}{2} + \frac{12}{2}$ $-2y_Q = \frac{9}{2}$

• Solve for $y_Q$ (the Ordinate of Q): $y_Q = \frac{9}{2} \cdot \left(-\frac{1}{2}\right)$ $y_Q = -\frac{9}{4}$

• Tip: Alternative Method for Q's Ordinate (useful property): For a parabola $y^2=4ax$, the tangent at $P(at^2, 2at)$ intersects the directrix $x=-a$ at the point $Q\left(-a, a\left(t - \frac{1}{t}\right)\right)$. Using this formula directly to verify our result: $y_Q = a\left(t - \frac{1}{t}\right)$ Substitute $a = \frac{3}{2}$ and $t = -2$: $y_Q = \frac{3}{2}\left(-2 - \frac{1}{-2}\right)$ $y_Q = \frac{3}{2}\left(-2 + \frac{1}{2}\right)$ $y_Q = \frac{3}{2}\left(-\frac{4}{2} + \frac{1}{2}\right)$ $y_Q = \frac{3}{2}\left(-\frac{3}{2}\right)$ $y_Q = -\frac{9}{4}$ Both methods yield the same result, confirming our calculation.

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Final Answer

The ordinate of the point Q is $-\frac{9}{4}$.

The final answer is $\boxed{\text{-}\frac{9}{4}}$.

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Summary and Key Takeaways

This problem is a comprehensive test of your understanding of parabola properties and algebraic manipulation.

1. Parametric Form is Essential: Using the parametric representation $P(at^2, 2at)$ is the most efficient way to handle problems involving tangents and normals to a parabola, simplifying calculations significantly.
2. Systematic Approach to Cubic Equations: Solving cubic