The area bounded by the normal at `(1, 2)` to the parabole `y^(2)=4x`, x-axis and the curve is …...........

Answer

To find the area bounded by the normal at the point (1, 2) to the parabola \(y^2 = 4x\), the x-axis, and the curve itself, we will follow these steps:

### Step 1: Find the slope of the tangent to the parabola at the point (1, 2).

The equation of the parabola is given by:
\[ y^2 = 4x \]

To find the slope of the tangent, we differentiate the equation with respect to \(x\):
\[ \frac{d(y^2)}{dx} = \frac{d(4x)}{dx} \implies 2y \frac{dy}{dx} = 4 \implies \frac{dy}{dx} = \frac{4}{2y} = \frac{2}{y} \]

At the point (1, 2), substituting \(y = 2\):
\[ \frac{dy}{dx} = \frac{2}{2} = 1 \]

### Step 2: Find the slope of the normal line.

The slope of the normal line is the negative reciprocal of the slope of the tangent:
\[ \text{slope of normal} = -\frac{1}{\frac{dy}{dx}} = -1 \]

### Step 3: Write the equation of the normal line.

Using the point-slope form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]
where \(m\) is the slope and \((x_1, y_1) = (1, 2)\):
\[ y - 2 = -1(x - 1) \implies y - 2 = -x + 1 \implies x + y = 3 \]

### Step 4: Find the points of intersection between the normal line and the parabola.

We need to find the intersection of the line \(x + y = 3\) with the parabola \(y^2 = 4x\). From the line equation, we can express \(y\):
\[ y = 3 - x \]

Substituting into the parabola equation:
\[ (3 - x)^2 = 4x \]
Expanding and rearranging:
\[ 9 - 6x + x^2 = 4x \implies x^2 - 10x + 9 = 0 \]

### Step 5: Solve the quadratic equation.

Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
\[ x = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 1 \cdot 9}}{2 \cdot 1} = \frac{10 \pm \sqrt{100 - 36}}{2} = \frac{10 \pm \sqrt{64}}{2} = \frac{10 \pm 8}{2} \]
Thus, the solutions are:
\[ x = \frac{18}{2} = 9 \quad \text{and} \quad x = \frac{2}{2} = 1 \]

### Step 6: Find the corresponding \(y\) values.

For \(x = 1\):
\[ y = 3 - 1 = 2 \]
For \(x = 9\):
\[ y = 3 - 9 = -6 \]

### Step 7: Calculate the area bounded by the curves.

We need to find the area between the parabola and the normal line from \(x = 1\) to \(x = 9\).

The area \(A\) can be computed as:
\[ A = \int_{1}^{9} \left( (3 - x) - \sqrt{4x} \right) \, dx \]

### Step 8: Evaluate the integral.

Calculating the integral:
1. The integral of \(3 - x\) from 1 to 9:
\[ \int_{1}^{9} (3 - x) \, dx = \left[ 3x - \frac{x^2}{2} \right]_{1}^{9} = \left( 27 - \frac{81}{2} \right) - \left( 3 - \frac{1}{2} \right) = \left( 27 - 40.5 \right) - \left( 3 - 0.5 \right) = -13.5 - 2.5 = -16 \]

2. The integral of \(\sqrt{4x} = 2\sqrt{x}\):
\[ \int_{1}^{9} 2\sqrt{x} \, dx = \left[ \frac{4}{3} x^{3/2} \right]_{1}^{9} = \frac{4}{3} \left( 27 - 1 \right) = \frac{4}{3} \cdot 26 = \frac{104}{3} \]

### Step 9: Combine the areas.

The total area \(A\) is:
\[ A = \left( -16 \right) - \left( -\frac{104}{3} \right) = -16 + \frac{104}{3} = \frac{-48 + 104}{3} = \frac{56}{3} \]

### Final Answer:
The area bounded by the normal at (1, 2) to the parabola \(y^2 = 4x\), the x-axis, and the curve is:
\[ \frac{56}{3} \text{ square units.} \]