To find the area of the triangle formed by the tangent and normal at the point (8, 8) on the parabola \( y^2 = 8x \) and the axis of the parabola, we will follow these steps:
### Step 1: Find the slope of the tangent at the point (8, 8).
The equation of the parabola is given by:
\[
y^2 = 8x
\]
Differentiating both sides with respect to \( x \):
\[
2y \frac{dy}{dx} = 8
\]
\[
\frac{dy}{dx} = \frac{8}{2y} = \frac{4}{y}
\]
Now, substitute \( y = 8 \) to find the slope at the point (8, 8):
\[
\frac{dy}{dx} = \frac{4}{8} = \frac{1}{2}
\]
### Step 2: Write the equation of the tangent line.
The equation of the tangent line at the point \( (x_0, y_0) \) is given by:
\[
y - y_0 = m(x - x_0)
\]
Substituting \( (x_0, y_0) = (8, 8) \) and \( m = \frac{1}{2} \):
\[
y - 8 = \frac{1}{2}(x - 8)
\]
Rearranging gives:
\[
y = \frac{1}{2}x + 4
\]
### Step 3: Find the x-intercept of the tangent line.
To find the x-intercept, set \( y = 0 \):
\[
0 = \frac{1}{2}x + 4
\]
\[
\frac{1}{2}x = -4 \implies x = -8
\]
Thus, the tangent line intersects the x-axis at the point \( (-8, 0) \).
### Step 4: Find the slope of the normal line.
The slope of the normal line is the negative reciprocal of the slope of the tangent:
\[
m_{\text{normal}} = -\frac{1}{\left(\frac{1}{2}\right)} = -2
\]
### Step 5: Write the equation of the normal line.
Using the point-slope form:
\[
y - 8 = -2(x - 8)
\]
Rearranging gives:
\[
y = -2x + 24
\]
### Step 6: Find the x-intercept of the normal line.
To find the x-intercept, set \( y = 0 \):
\[
0 = -2x + 24
\]
\[
2x = 24 \implies x = 12
\]
Thus, the normal line intersects the x-axis at the point \( (12, 0) \).
### Step 7: Determine the vertices of the triangle.
The vertices of the triangle formed by the tangent, normal, and the x-axis are:
1. \( (-8, 0) \) (x-intercept of the tangent)
2. \( (12, 0) \) (x-intercept of the normal)
3. \( (8, 8) \) (the point of tangency)
### Step 8: Calculate the area of the triangle.
The base of the triangle is the distance between the x-intercepts:
\[
\text{Base} = 12 - (-8) = 12 + 8 = 20
\]
The height of the triangle is the y-coordinate of the point of tangency:
\[
\text{Height} = 8
\]
The area \( A \) of the triangle is given by:
\[
A = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 20 \times 8 = 80 \text{ square units}
\]
### Final Answer:
The area of the triangle is \( 80 \) square units.
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