This solution will guide you through a comprehensive approach to solving problems involving parabolas, specifically focusing on finding the equation of a parabola, determining intersection points, and calculating the distance between points related by a normal. We will break down each step, explaining the underlying concepts and calculations, and highlight useful formulas and common pitfalls.

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1. Understanding the Parabola's Properties and Deriving its Equation

The fundamental definition of a parabola is the locus of all points that are equidistant from a fixed point (the focus) and a fixed line (the directrix).

Given:

• Vertex $V = \left( {{1 \over 2},{3 \over 4}} \right) = (h, k)$
• Directrix $y = {1 \over 2}$

Since the directrix is a horizontal line ($y = \text{constant}$), the axis of symmetry must be a vertical line ($x = \text{constant}$). This means the parabola opens either upwards or downwards. The standard equation for such a parabola is: $(x-h)^2 = 4a(y-k)$ Here, $a$ represents the distance from the vertex to the focus, and also the distance from the vertex to the directrix.

• The directrix for this form is given by $y = k-a$.
• The focus for this form is given by $(h, k+a)$.

Let's use the given information to find the value of $a$: The directrix is $y = k-a$. Substituting the given vertex coordinates: $y = \frac{3}{4} - a$ We are given that the directrix is $y = \frac{1}{2}$. Therefore, $\frac{1}{2} = \frac{3}{4} - a$ Now, solve for $a$: $a = \frac{3}{4} - \frac{1}{2} = \frac{3}{4} - \frac{2}{4} = \frac{1}{4}$ Since $a > 0$, the parabola opens upwards.

Now, substitute $h = \frac{1}{2}$, $k = \frac{3}{4}$, and $a = \frac{1}{4}$ into the standard equation: $\left( x - \frac{1}{2} \right)^2 = 4 \left( \frac{1}{4} \right) \left( y - \frac{3}{4} \right)$ $\left( x - \frac{1}{2} \right)^2 = 1 \cdot \left( y - \frac{3}{4} \right)$ $\left( x - \frac{1}{2} \right)^2 = y - \frac{3}{4}$ This is the equation of the parabola.

Why this step is taken: Establishing the correct equation of the parabola is crucial as all subsequent calculations (finding points, tangents, normals) depend on it. Understanding the orientation and key parameters like 'a' from the vertex and directrix is a fundamental skill in parabola problems.

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2. Finding the Coordinates of Point P

Point P is where the parabola meets the line $x = - \frac{1}{2}$. To find its coordinates, substitute $x = - \frac{1}{2}$ into the parabola's equation: $\left( - \frac{1}{2} - \frac{1}{2} \right)^2 = y - \frac{3}{4}$ $(-1)^2 = y - \frac{3}{4}$ $1 = y - \frac{3}{4}$ Solve for $y$: $y = 1 + \frac{3}{4} = \frac{4}{4} + \frac{3}{4} = \frac{7}{4}$ So, the coordinates of point P are $\left( - \frac{1}{2}, \frac{7}{4} \right)$.

Why this step is taken: Point P is the specific point on the parabola where the normal is drawn. Its coordinates are essential for calculating the normal's equation and for the final distance calculation.

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3. Determining the Equation of the Normal at P and Finding Point Q

To find the normal, it's often convenient to transform the coordinate system such that the vertex is at the origin. Let: $X = x - h = x - \frac{1}{2}$ $Y = y - k = y - \frac{3}{4}$ The equation of the parabola becomes $X^2 = 4aY$. In our case, $a = \frac{1}{4}$, so: $X^2 = 4 \left( \frac{1}{4} \right) Y \implies X^2 = Y$ Now, let's find the coordinates of P in this new $(X, Y)$ system: $X_P = x_P - \frac{1}{2} = - \frac{1}{2} - \frac{1}{2} = -1$ $Y_P = y_P - \frac{3}{4} = \frac{7}{4} - \frac{3}{4} = \frac{4}{4} = 1$ So, $P = (-1, 1)$ in the $(X, Y)$ system.

For a parabola $X^2 = 4aY$, the slope of the tangent at a point $(X_1, Y_1)$ is found by differentiating with respect to $X$: $2X = 4a \frac{dY}{dX} \implies \frac{dY}{dX} = \frac{2X}{4a} = \frac{X}{2a}$ At point $P(X_P, Y_P)$, the slope of the tangent is $m_T = \frac{X_P}{2a}$. The slope of the normal, $m_N$, is the negative reciprocal of the tangent's slope: $m_N = - \frac{2a}{X_P}$ Substituting $X_P = -1$ and $a = \frac{1}{4}$: $m_N = - \frac{2 \left( \frac{1}{4} \right)}{-1} = - \frac{\frac{1}{2}}{-1} = \frac{1}{2}$ The equation of the normal at $P(X_P, Y_P)$ in the $(X, Y)$ system is: $Y - Y_P = m_N (X - X_P)$ $Y - 1 = \frac{1}{2} (X - (-1))$ $Y - 1 = \frac{1}{2} (X + 1)$

Now, to find point Q, where this normal intersects the parabola again, we substitute $Y = X^2$ (from $X^2 = Y$) into the normal's equation: $X^2 - 1 = \frac{1}{2} (X + 1)$ Multiply by 2: $2(X^2 - 1) = X + 1$ $2(X-1)(X+1) = X + 1$ We know that $X = X_P = -1$ is one solution (corresponding to point P). Since $X+1$ is a common factor, we can divide by it, assuming $X \ne -1$ (which holds for Q, as Q is a different intersection point). $2(X-1) = 1$ $2X - 2 = 1$ $2X = 3$ $X_Q = \frac{3}{2}$ Now find $Y_Q$ using $Y_Q = X_Q^2$: $Y_Q = \left( \frac{3}{2} \right)^2 = \frac{9}{4}$ So, the coordinates of Q in the $(X, Y)$ system are $\left( \frac{3}{2}, \frac{9}{4} \right)$.

Tip for finding the second intersection of a normal: For a parabola of the form $X^2 = 4aY$, if the normal at $P(X_P, Y_P)$ intersects the parabola again at