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The given problem involves finding the point of intersection of the tangents to the graph at points \( P \) and \( Q \), both of which lie on the curve described by the equation:

\[
y^2 - xy + 8 = 0
\]

It is given that the points \( P \) and \( Q \) have the same x-coordinate, \( x = 6 \). The options provided give coordinates for the intersection of the tangents at these points.

### Step 1: Differentiate the equation implicitly

To find the slopes of the tangents at points \( P \) and \( Q \), we need to find the derivative of \( y^2 - xy + 8 = 0 \) with respect to \( x \). Using implicit differentiation:

\[
\frac{d}{dx}(y^2 - xy + 8) = 0
\]

Applying the product rule to \( xy \), we get:

\[
2y \frac{dy}{dx} - \left( x \frac{dy}{dx} + y \right) = 0
\]

Simplifying:

\[
2y \frac{dy}{dx} - x \frac{dy}{dx} - y = 0
\]

\[
(2y - x) \frac{dy}{dx} = y
\]

Thus, the slope of the tangent line (i.e., \( \frac{dy}{dx} \)) is:

\[
\frac{dy}{dx} = \frac{y}{2y - x}
\]

### Step 2: Find the y-coordinates for \( P \) and \( Q \)

Since \( P \) and \( Q \) share the same x-coordinate, \( x = 6 \), substitute \( x = 6 \) into the original equation to find the possible y-values:

\[
y^2 - 6y + 8 = 0
\]

Solve this quadratic equation:

\[
y = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(8)}}{2(1)}
\]

\[
y = \frac{6 \pm \sqrt{36 - 32}}{2}
\]

\[
y = \frac{6 \pm \sqrt{4}}{2}
\]

\[
y = \frac{6 \pm 2}{2}
\]

Thus, the two solutions for \( y \) are:

\[
y = 4 \quad \text{and} \quad y = 2
\]

So, the points \( P \) and \( Q \) are \( (6, 4) \) and \( (6, 2) \), respectively.

### Step 3: Calculate the slopes at \( P \) and \( Q \)

Using the formula for the slope \( \frac{dy}{dx} = \frac{y}{2y - x} \), substitute \( x = 6 \) and the y-values:

- At \( P(6, 4) \), the slope is:

\[
\frac{dy}{dx} = \frac{4}{2(4) - 6} = \frac{4}{8 - 6} = \frac{4}{2} = 2
\]

- At \( Q(6, 2) \), the slope is:

\[
\frac{dy}{dx} = \frac{2}{2(2) - 6} = \frac{2}{4 - 6} = \frac{2}{-2} = -1
\]

### Step 4: Find the equations of the tangent lines at \( P \) and \( Q \)

- The tangent line at \( P(6, 4) \) with slope \( 2 \) is:

\[
y - 4 = 2(x - 6)
\]

Simplifying:

\[
y = 2x - 12 + 4 = 2x - 8
\]

- The tangent line at \( Q(6, 2) \) with slope \( -1 \) is:

\[
y - 2 = -1(x - 6)
\]

Simplifying:

\[
y = -x + 6 + 2 = -x + 8
\]

### Step 5: Solve for the intersection of the tangents

To find the intersection of the tangents, set the equations equal to each other:

\[
2x - 8 = -x + 8
\]

Solving for \( x \):

\[
2x + x = 8 + 8
\]

\[
3x = 16
\]

\[
x = \frac{16}{3}
\]

Substitute \( x = \frac{16}{3} \) into either equation (we'll use \( y = 2x - 8 \)):

\[
y = 2\left(\frac{16}{3}\right) - 8 = \frac{32}{3} - 8 = \frac{32}{3} - \frac{24}{3} = \frac{8}{3}
\]

Thus, the point of intersection of the tangents is:

\[
\left( \frac{16}{3}, \frac{8}{3} \right)
\]

### Answer:
The correct answer is \( \boxed{(C)} \).

Would you like more details on any part of the solution? Here are 5 related questions to deepen your understanding:

1. How is implicit differentiation used in solving problems involving curves and tangents?
2. What are the properties of the quadratic equation, and how can they be applied to solve for roots?
3. How do tangent lines relate to derivatives of a curve at a point?
4. What methods can be used to find the intersection of two linear equations?
5. Why is it important to understand the slope-intercept form of a line when solving tangent problems?

**Tip:** When differentiating implicitly, always apply the chain rule carefully, especially when dealing with products or higher-order terms.

The points P and Q on the graph of y^2 - xy + 8 = 0 have the same x-coordinate, x = 6. The point of intersection of the tangents to the graph at P and Q is?

Explore how to find the intersection of tangents to the curve y^2 - xy + 8 = 0 at points P and Q where x = 6. This solution uses implicit differentiation, quadratic equations, and linear equation intersections. Suitable for Grades 11-12.

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