The line `2y=3x+12` cuts the parabola `4y=3x^(2)`
What is the area enclosed by the parabola, the line and the y-axis in the first quadrant?

Options

A. 7 square unit
B. 14 square unit
C. 20 square unit
D. 21 square unit

Answer

To find the area enclosed by the parabola \(4y = 3x^2\), the line \(2y = 3x + 12\), and the y-axis in the first quadrant, we will follow these steps:

### Step 1: Convert the equations to standard form
First, let's rewrite the equations in terms of \(y\):
- For the parabola:
\[ y = \frac{3}{4}x^2 \]
- For the line:
\[ y = \frac{3}{2}x + 6 \]

### Step 2: Find the points of intersection
To find the points where the line intersects the parabola, we set the two equations for \(y\) equal to each other:
\[ \frac{3}{4}x^2 = \frac{3}{2}x + 6 \]
Multiplying through by 4 to eliminate the fraction:
\[ 3x^2 = 6x + 24 \]
Rearranging gives:
\[ 3x^2 - 6x - 24 = 0 \]
Dividing the entire equation by 3:
\[ x^2 - 2x - 8 = 0 \]
Factoring the quadratic:
\[ (x - 4)(x + 2) = 0 \]
Thus, the solutions for \(x\) are:
\[ x = 4 \quad \text{and} \quad x = -2 \]
Since we are interested in the first quadrant, we take \(x = 4\).

### Step 3: Find the corresponding \(y\) values
Now we substitute \(x = 4\) back into either equation to find \(y\):
Using the line equation:
\[ y = \frac{3}{2}(4) + 6 = 6 + 6 = 12 \]
So the point of intersection in the first quadrant is \((4, 12)\).

### Step 4: Set up the integral for the area
The area \(A\) between the line and the parabola from \(x = 0\) to \(x = 4\) can be found using the integral:
\[ A = \int_0^4 \left( \text{(line)} - \text{(parabola)} \right) \, dx \]
Substituting the equations:
\[ A = \int_0^4 \left( \left(\frac{3}{2}x + 6\right) - \left(\frac{3}{4}x^2\right) \right) \, dx \]

### Step 5: Simplify the integral
This simplifies to:
\[ A = \int_0^4 \left( \frac{3}{2}x + 6 - \frac{3}{4}x^2 \right) \, dx \]

### Step 6: Calculate the integral
Now we calculate the integral:
\[ A = \int_0^4 \left( \frac{3}{2}x + 6 - \frac{3}{4}x^2 \right) \, dx \]
Calculating the integral term by term:
\[ = \left[\frac{3}{4}x^2 + 6x - \frac{3}{12}x^3 \right]_0^4 \]
Evaluating at the limits:
\[ = \left[\frac{3}{4}(4^2) + 6(4) - \frac{3}{12}(4^3)\right] - \left[\frac{3}{4}(0^2) + 6(0) - \frac{3}{12}(0^3)\right] \]
Calculating each term:
\[ = \left[\frac{3}{4}(16) + 24 - \frac{3}{12}(64)\right] \]
\[ = \left[12 + 24 - 16\right] \]
\[ = 20 \]

### Final Answer
Thus, the area enclosed by the parabola, the line, and the y-axis in the first quadrant is \(20\) square units.

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