Let B and C are points of interection of the parabola `y=x^(2)` and the circle `x^(2)+(y-2)^(2)=8.` The area of the triangle OBC, where O is the origin, is

Options

A. 2
B. 4
C. 6
D. 8

Answer

To find the area of triangle OBC formed by the points of intersection of the parabola \( y = x^2 \) and the circle \( x^2 + (y - 2)^2 = 8 \), we will follow these steps:

### Step 1: Find the points of intersection
We start with the equations of the parabola and the circle:
- Parabola: \( y = x^2 \)
- Circle: \( x^2 + (y - 2)^2 = 8 \)

Substituting \( y = x^2 \) into the circle's equation:
\[ x^2 + (x^2 - 2)^2 = 8 \]
Expanding the equation:
\[ x^2 + (x^4 - 4x^2 + 4) = 8 \]
This simplifies to:
\[ x^4 - 3x^2 - 4 = 0 \]

### Step 2: Solve the quadratic equation
Let \( z = x^2 \). Then we have:
\[ z^2 - 3z - 4 = 0 \]
Using the quadratic formula \( z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ z = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 1 \cdot (-4)}}{2 \cdot 1} \]
\[ z = \frac{3 \pm \sqrt{9 + 16}}{2} \]
\[ z = \frac{3 \pm 5}{2} \]
Calculating the two possible values:
\[ z = 4 \quad \text{or} \quad z = -1 \]
Since \( z = x^2 \), we discard \( z = -1 \) as it is not valid. Thus, \( z = 4 \) gives:
\[ x^2 = 4 \implies x = \pm 2 \]

### Step 3: Find the corresponding y-coordinates
Substituting \( x = 2 \) and \( x = -2 \) back into the parabola's equation:
\[ y = x^2 = 4 \]
Thus, the points of intersection are:
- \( B(-2, 4) \)
- \( C(2, 4) \)

### Step 4: Calculate the area of triangle OBC
The coordinates of the points are:
- \( O(0, 0) \)
- \( B(-2, 4) \)
- \( C(2, 4) \)

Using the formula for the area of a triangle formed by points \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\):
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
Substituting the coordinates:
\[ \text{Area} = \frac{1}{2} \left| 0(4 - 4) + (-2)(4 - 0) + 2(0 - 4) \right| \]
\[ = \frac{1}{2} \left| 0 - 8 - 8 \right| = \frac{1}{2} \left| -16 \right| = \frac{16}{2} = 8 \]

### Final Answer
The area of triangle OBC is \( 8 \) square units.

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