The line `2y=3x+12` cuts the parabola `4y=3x^(2)`
Where does the line cut the parabola ?

Options

A. A. At (-2,3) only
B. B. At (4,12)only
C. C. At both (-2,3) and (4,12)
D. D. Neither at (-2,3) nor at (4,12)

Answer

To find the points where the line \(2y = 3x + 12\) intersects the parabola \(4y = 3x^2\), we will follow these steps:

### Step 1: Write down the equations
The equations given are:
1. Line: \(2y = 3x + 12\)
2. Parabola: \(4y = 3x^2\)

### Step 2: Express \(y\) in terms of \(x\)
From the equation of the parabola, we can express \(y\) in terms of \(x\):
\[ y = \frac{3x^2}{4} \]

### Step 3: Substitute \(y\) into the line's equation
Now, substitute \(y\) from the parabola into the line's equation:
\[ 2\left(\frac{3x^2}{4}\right) = 3x + 12 \]

### Step 4: Simplify the equation
Multiply both sides by 4 to eliminate the fraction:
\[ 2 \cdot 3x^2 = 4(3x + 12) \]
This simplifies to:
\[ 6x^2 = 12x + 48 \]

### Step 5: Rearrange the equation
Rearranging gives us:
\[ 6x^2 - 12x - 48 = 0 \]
Dividing the entire equation by 6 simplifies it further:
\[ x^2 - 2x - 8 = 0 \]

### Step 6: Factor the quadratic equation
We can factor the quadratic:
\[ (x - 4)(x + 2) = 0 \]
Thus, the solutions for \(x\) are:
\[ x = 4 \quad \text{and} \quad x = -2 \]

### Step 7: Find corresponding \(y\) values
Now, we will find the corresponding \(y\) values for both \(x\) values using the equation of the parabola \(y = \frac{3x^2}{4}\).

1. For \(x = 4\):
\[ y = \frac{3(4^2)}{4} = \frac{3 \cdot 16}{4} = 12 \]
So, one point of intersection is \((4, 12)\).

2. For \(x = -2\):
\[ y = \frac{3(-2)^2}{4} = \frac{3 \cdot 4}{4} = 3 \]
So, the other point of intersection is \((-2, 3)\).

### Step 8: Conclusion
The points where the line cuts the parabola are:
\[ (-2, 3) \quad \text{and} \quad (4, 12) \]

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