The parabola `y=4-x^(2)` has vertex P. It intersects x-axis at A and B. If the parabola is translated from its initial position to a new position by moving its vertex along the line `y=x+4`, so that it intersects x-axis at B and C, then abscissa of C will be :

Options

A. 3
B. 4
C. 5
D. 8

Answer

To solve the problem step by step, we will follow the reasoning presented in the video transcript.

### Step 1: Identify the original parabola and its vertex
The original parabola is given by the equation:
\[ y = 4 - x^2 \]
The vertex of this parabola is at the point \( P(0, 4) \).

### Step 2: Find the intersection points with the x-axis
To find the intersection points with the x-axis, we set \( y = 0 \):
\[ 0 = 4 - x^2 \]
This simplifies to:
\[ x^2 = 4 \]
Taking the square root gives:
\[ x = 2 \quad \text{or} \quad x = -2 \]
Thus, the points of intersection are:
\[ A(-2, 0) \quad \text{and} \quad B(2, 0) \]

### Step 3: Translate the vertex along the line \( y = x + 4 \)
We need to translate the vertex along the line \( y = x + 4 \). Let the new vertex be at point \( (\alpha, \alpha + 4) \).

### Step 4: Write the equation of the new parabola
The equation of the new parabola can be expressed in vertex form as:
\[ (y - (\alpha + 4)) = -(x - \alpha)^2 \]
Rearranging gives:
\[ (x - \alpha)^2 = -(y - (\alpha + 4)) \]

### Step 5: Substitute the known point \( B(2, 0) \)
Since the new parabola must pass through point \( B(2, 0) \), we substitute \( x = 2 \) and \( y = 0 \) into the equation:
\[ (2 - \alpha)^2 = - (0 - (\alpha + 4)) \]
This simplifies to:
\[ (2 - \alpha)^2 = \alpha + 4 \]

### Step 6: Expand and rearrange the equation
Expanding the left side:
\[ 4 - 4\alpha + \alpha^2 = \alpha + 4 \]
Rearranging gives:
\[ \alpha^2 - 5\alpha = 0 \]

### Step 7: Factor the equation
Factoring out \( \alpha \):
\[ \alpha(\alpha - 5) = 0 \]
This gives us two possible solutions:
\[ \alpha = 0 \quad \text{or} \quad \alpha = 5 \]
Since the vertex cannot be at the original position, we take:
\[ \alpha = 5 \]

### Step 8: Write the equation of the new parabola
Substituting \( \alpha = 5 \) back into the vertex form gives:
\[ (x - 5)^2 = -(y - 9) \]
or
\[ (x - 5)^2 = -y + 9 \]
Thus, the new parabola is:
\[ y = - (x - 5)^2 + 9 \]

### Step 9: Find the intersection points with the x-axis
To find the x-intercepts (where \( y = 0 \)):
\[ 0 = - (x - 5)^2 + 9 \]
Rearranging gives:
\[ (x - 5)^2 = 9 \]

### Step 10: Solve for \( x \)
Taking the square root:
\[ x - 5 = 3 \quad \text{or} \quad x - 5 = -3 \]
Thus:
\[ x = 8 \quad \text{or} \quad x = 2 \]
Since \( x = 2 \) corresponds to point \( B \), the other point \( C \) has an abscissa of:
\[ x = 8 \]

### Final Answer
The abscissa of point \( C \) is:
\[ \boxed{8} \]