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Let's tackle the problems step by step.

### Question 27
Given:
- Linear function: \( y = \frac{3}{4}x \)
- Quadratic function: \( y = ax^2 - 4ax + c \)

#### Part (1):
Find the coordinates of point \( C \) where the two graphs intersect.

The intersection points are solutions to:
\[ \frac{3}{4}x = ax^2 - 4ax + c \]

#### Part (2):
Assume the vertex of the quadratic function is point \( D \).

**(i)** If point \( D \) is symmetric about the x-axis with point \( C \), and the area of triangle \( \Delta ACD \) is 3, find the relationship of the quadratic function.

**(ii)** If \( CD = AC \), and the area of triangle \( \Delta ACD \) is 10, find the relationship of the quadratic function.

### Question 28
Given:
- \( \angle AOB = 60^\circ \)
- \( OM = 4 \)
- \( OQ = 1 \)

#### Part (1):
Prove \( CN \perp OB \).

#### Part (2):
When point \( N \) moves along \( OB \), the quadrilateral \( OMPQ \) remains a parallelogram.

**(i)** Does the value of \( \frac{1}{OM} - \frac{1}{ON} \) change? If it does, find the range; if not, explain why.

### Solution Steps:

#### Question 27 Solution:
1. **Finding Point \( C \):**
   Solve \( \frac{3}{4}x = ax^2 - 4ax + c \):
   \[ ax^2 - 4ax - \frac{3}{4}x + c = 0 \]
   \[ ax^2 - \left(4 + \frac{3}{4}\right)x + c = 0 \]
   \[ ax^2 - \frac{19}{4}x + c = 0 \]
   Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) to find \( x \).

2. **Vertex \( D \) and Symmetry:**
   - For \( D \) symmetric about the x-axis with \( C \):
     If \( D \) is the vertex, \( x_D = -x_C \).
     Calculate coordinates and area to find \( a, c \).

3. **Area Condition:**
   - For the area of \( \Delta ACD \) to be 3 or 10, use the given conditions to establish relationships between \( a \) and \( c \).

#### Question 28 Solution:
1. **Proving \( CN \perp OB \):**
   - Calculate slopes of \( CN \) and \( OB \) and show they are negative reciprocals.

2. **Quadrilateral \( OMPQ \):**
   - Show properties of a parallelogram.
   - Check if \( \frac{1}{OM} - \frac{1}{ON} \) changes as \( N \) moves.

Let's start with the intersection calculation for Question 27.

### Calculation:
For the intersection:
\[ \frac{3}{4}x = ax^2 - 4ax + c \]
\[ ax^2 - \frac{19}{4}x + c = 0 \]

Using the quadratic formula \( x = \frac{\frac{19}{4} \pm \sqrt{(\frac{19}{4})^2 - 4ac}}{2a} \):

\[ x = \frac{\frac{19}{4} \pm \sqrt{\frac{361}{16} - 4ac}}{2a} \]
\[ x = \frac{\frac{19}{4} \pm \sqrt{\frac{361}{16} - 4ac}}{2a} \]

Let's handle these calculations step by step if you'd like to proceed or any particular section to detail more, please let me know!

Would you like to continue with these calculations, or have any questions on a specific part? 

### Here are five related questions you might consider next:
1. How to find the vertex form of a quadratic equation?
2. What is the significance of the discriminant in a quadratic equation?
3. How to determine the area of a triangle given vertex coordinates?
4. How to prove two lines are perpendicular using slopes?
5. What are the properties of a parallelogram in coordinate geometry?

### Tip:
When solving quadratic equations, always ensure to simplify coefficients where possible to make calculations easier.

Explore the detailed solution for finding the intersection of a linear function and a quadratic function, along with proving geometric properties. Ideal for high school students studying algebra and geometry.

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