To solve the problem, we need to find the area of triangle ATP formed by the points A(2, 2), T, and P, where T is the point where the tangent to the curve \(y^2 = 2x\) at point A intersects the x-axis, and P is the point where the normal at A intersects the curve again.
### Step-by-Step Solution:
1. **Find the Tangent Line at Point A(2, 2)**:
The equation of the parabola is given by \(y^2 = 2x\). The slope of the tangent at any point \((x_1, y_1)\) on the parabola can be found using the derivative.
The derivative of \(y^2 = 2x\) with respect to \(x\) is:
\[
2y \frac{dy}{dx} = 2 \implies \frac{dy}{dx} = \frac{1}{y}
\]
At point A(2, 2), the slope \(m\) of the tangent is:
\[
m = \frac{1}{2}
\]
The equation of the tangent line at point A can be written using the point-slope form:
\[
y - y_1 = m(x - x_1) \implies y - 2 = \frac{1}{2}(x - 2)
\]
Simplifying this gives:
\[
2y - 4 = x - 2 \implies 2y = x + 2 \implies x - 2y + 2 = 0
\]
2. **Find the Point T where the Tangent Cuts the X-axis**:
To find the x-intercept (point T), set \(y = 0\) in the tangent equation:
\[
x - 2(0) + 2 = 0 \implies x + 2 = 0 \implies x = -2
\]
Therefore, point T is \((-2, 0)\).
3. **Find the Normal Line at Point A(2, 2)**:
The slope of the normal line is the negative reciprocal of the tangent slope. Thus, the slope of the normal is:
\[
m_{normal} = -2
\]
Using the point-slope form for the normal line:
\[
y - 2 = -2(x - 2)
\]
Simplifying this gives:
\[
y - 2 = -2x + 4 \implies y = -2x + 6
\]
4. **Find the Point P where the Normal Intersects the Curve Again**:
Substitute \(y = -2x + 6\) into the parabola equation \(y^2 = 2x\):
\[
(-2x + 6)^2 = 2x
\]
Expanding this:
\[
4x^2 - 24x + 36 = 2x \implies 4x^2 - 26x + 36 = 0
\]
Dividing through by 2:
\[
2x^2 - 13x + 18 = 0
\]
Using the quadratic formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{13 \pm \sqrt{(-13)^2 - 4 \cdot 2 \cdot 18}}{2 \cdot 2}
\]
\[
= \frac{13 \pm \sqrt{169 - 144}}{4} = \frac{13 \pm 5}{4}
\]
This gives:
\[
x = \frac{18}{4} = \frac{9}{2} \quad \text{and} \quad x = \frac{8}{4} = 2
\]
Since \(x = 2\) corresponds to point A, we take \(x = \frac{9}{2}\).
Now, substituting \(x = \frac{9}{2}\) back into the normal equation to find \(y\):
\[
y = -2\left(\frac{9}{2}\right) + 6 = -9 + 6 = -3
\]
Thus, point P is \(\left(\frac{9}{2}, -3\right)\).
5. **Calculate the Area of Triangle ATP**:
The area of triangle formed by points A(2, 2), T(-2, 0), and P\(\left(\frac{9}{2}, -3\right)\) can be calculated using the determinant formula:
\[
\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|
\]
Substituting the coordinates:
\[
= \frac{1}{2} \left| 2(0 - (-3)) + (-2)(-3 - 2) + \frac{9}{2}(2 - 0) \right|
\]
\[
= \frac{1}{2} \left| 2(3) + (-2)(-5) + \frac{9}{2}(2) \right|
\]
\[
= \frac{1}{2} \left| 6 + 10 + 9 \right| = \frac{1}{2} \left| 25 \right| = \frac{25}{2}
\]
Thus, the area of triangle ATP is \(\frac{25}{2}\) square units.
