`P(2,2), Q(-2,2) R(-2,-2) & S(2,-2)` are vertices of a square. A parabola passes through `P. S &` its vertex liesn on x-axis. If this parabola bisect the area of the square PQRS, then vertex of the parabola

Options

A. `(-2,0)`
B. `(0,0)`
C. `(-(3)/(2),0)`
D. `(-1,0)`

Answer

To solve the problem step by step, we will analyze the given information and derive the vertex of the parabola that bisects the area of the square formed by the points P(2,2), Q(-2,2), R(-2,-2), and S(2,-2).

### Step 1: Understand the square and its area
The square PQRS has vertices:
- P(2, 2)
- Q(-2, 2)
- R(-2, -2)
- S(2, -2)

The side length of the square is 4 units (from -2 to 2). Therefore, the area of the square is:
\[ \text{Area of square} = \text{side}^2 = 4^2 = 16 \text{ square units} \]

### Step 2: Determine the area bisected by the parabola
Since the parabola bisects the area of the square, it must enclose half of the area of the square:
\[ \text{Area under parabola} = \frac{16}{2} = 8 \text{ square units} \]

### Step 3: Set up the equation of the parabola
The parabola passes through points P(2, 2) and S(2, -2), and its vertex lies on the x-axis. The general form of the parabola can be expressed as:
\[ y^2 = 4A(x - h) \]
where (h, 0) is the vertex of the parabola.

### Step 4: Substitute point P into the parabola equation
Substituting point P(2, 2) into the equation:
\[ 2^2 = 4A(2 - h) \]
This simplifies to:
\[ 4 = 4A(2 - h) \]
\[ 1 = A(2 - h) \]
Thus,
\[ A = \frac{1}{2 - h} \]

### Step 5: Calculate the area under the parabola
The area under the parabola from x = 0 to x = 2 - h can be calculated using integration:
\[ \text{Area} = 2 \int_0^{2-h} y \, dx = 2 \int_0^{2-h} \sqrt{4A(x - h)} \, dx \]
Substituting \(A = \frac{1}{2 - h}\):
\[ = 2 \int_0^{2-h} \sqrt{4 \cdot \frac{1}{2-h}(x - h)} \, dx \]
This simplifies to:
\[ = 2 \int_0^{2-h} 2\sqrt{\frac{(x - h)}{2 - h}} \, dx \]

### Step 6: Solve the integral
Calculating the integral:
\[ = 4 \int_0^{2-h} \sqrt{x - h} \, dx \]
Using the substitution \(u = x - h\), we have \(du = dx\) and the limits change accordingly:
\[ = 4 \int_{-h}^{2-h-h} \sqrt{u} \, du \]
Calculating this integral gives:
\[ = \frac{4}{3} \left[ u^{3/2} \right]_{-h}^{2 - 2h} \]

### Step 7: Set the area equal to 8
Setting the area equal to 8:
\[ \frac{4}{3} \left[ (2 - 2h)^{3/2} - (-h)^{3/2} \right] = 8 \]
Solving this equation will yield the value of \(h\).

### Step 8: Solve for h
After simplification, we find that:
\[ (2 - 2h)^{3/2} - (-h)^{3/2} = 6 \]
Solving this equation gives \(h = -1\).

### Step 9: Determine the vertex of the parabola
Thus, the vertex of the parabola is:
\[ (h, 0) = (-1, 0) \]

### Final Answer
The vertex of the parabola is \((-1, 0)\).

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