The area between the parabola `y^2 = 4x` , normal at one end of latusrectum and X-axis in sq.units is

Options

A. `60^(@)`
B. less then `60^(@)`
C. more then`60^(@)`
D. less then `45^(@)`

Explanation

To find the area between the parabola \( y^2 = 4x \), the normal at one end of the latus rectum, and the x-axis, we can follow these steps:

### Step 1: Identify the parameters of the parabola
The given parabola is \( y^2 = 4x \). Here, we can identify \( a = 1 \) because it is in the form \( y^2 = 4ax \).

**Hint:** The value of \( a \) helps in determining the coordinates of the latus rectum.

### Step 2: Find the coordinates of the latus rectum
The latus rectum of the parabola \( y^2 = 4ax \) has endpoints at \( (a, 2a) \) and \( (a, -2a) \). For \( a = 1 \), the coordinates are:
- \( (1, 2) \)
- \( (1, -2) \)

**Hint:** The latus rectum is a vertical line segment through the focus of the parabola.

### Step 3: Determine the normal at one end of the latus rectum
We will find the normal at the point \( (1, 2) \). The slope \( m \) of the tangent at this point can be calculated using the derivative of the parabola.

From the equation \( y^2 = 4x \), we differentiate implicitly:
\[ 2y \frac{dy}{dx} = 4 \]
Thus,
\[ \frac{dy}{dx} = \frac{2}{y} \]
At the point \( (1, 2) \),
\[ \frac{dy}{dx} = \frac{2}{2} = 1 \]
The slope of the normal is the negative reciprocal:
\[ m = -1 \]

**Hint:** The slope of the normal line is perpendicular to the slope of the tangent line.

### Step 4: Write the equation of the normal line
Using the point-slope form of the line equation:
\[ y - y_1 = m(x - x_1) \]
Substituting \( (x_1, y_1) = (1, 2) \) and \( m = -1 \):
\[ y - 2 = -1(x - 1) \]
This simplifies to:
\[ y = -x + 3 \]

**Hint:** The normal line intersects the x-axis where \( y = 0 \).

### Step 5: Find the x-intercept of the normal line
Setting \( y = 0 \) in the equation of the normal:
\[ 0 = -x + 3 \]
Thus,
\[ x = 3 \]
So the x-intercept is \( (3, 0) \).

**Hint:** The x-intercept is where the normal line meets the x-axis.

### Step 6: Calculate the area between the parabola, the normal, and the x-axis
The area can be divided into two parts:
1. The area under the parabola from \( x = 0 \) to \( x = 1 \).
2. The area of the triangle formed by the points \( (1, 0) \), \( (3, 0) \), and \( (3, 2) \).

**Area under the parabola:**
To find the area under the parabola from \( x = 0 \) to \( x = 1 \):
\[ \text{Area}_\text{parabola} = \int_0^1 \sqrt{4x} \, dx = \int_0^1 2\sqrt{x} \, dx \]
Calculating the integral:
\[ = 2 \cdot \left[ \frac{2}{3} x^{3/2} \right]_0^1 = 2 \cdot \frac{2}{3} = \frac{4}{3} \]

**Area of the triangle:**
The base of the triangle is from \( x = 1 \) to \( x = 3 \) (length = 2) and the height is 2 (from \( (3, 0) \) to \( (3, 2) \)):
\[ \text{Area}_\text{triangle} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 2 \times 2 = 2 \]

### Step 7: Total area
Now, we add both areas:
\[ \text{Total Area} = \text{Area}_\text{parabola} + \text{Area}_\text{triangle} = \frac{4}{3} + 2 = \frac{4}{3} + \frac{6}{3} = \frac{10}{3} \]

**Final Answer:** The area between the parabola \( y^2 = 4x \), the normal at one end of the latus rectum, and the x-axis is \( \frac{10}{3} \) square units.

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