The straight line `(x)/(4)+(y)/(3) =1` intersects the ellipse `(x^(2))/(16)+(y^(2))/(9) =1` at two points A and B, there is a point P on this ellipse such that the area of `DeltaPAB` is equal to `6(sqrt(2)-1)`. Then the number of such points (P) is/are

Options

A. 0
B. 1
C. 2
D. 3

Explanation

To solve the problem step by step, we will analyze the given equations and find the number of points \( P \) on the ellipse such that the area of triangle \( PAB \) is equal to \( 6(\sqrt{2}-1) \).

### Step 1: Identify the equations of the line and the ellipse
The equations given are:
1. Line: \(\frac{x}{4} + \frac{y}{3} = 1\)
2. Ellipse: \(\frac{x^2}{16} + \frac{y^2}{9} = 1\)

### Step 2: Convert the line equation to standard form
Multiply through by 12 to eliminate the fractions:
\[ 3x + 4y = 12 \]
This can be rewritten as:
\[ 3x + 4y - 12 = 0 \]

### Step 3: Find the points of intersection A and B
To find the points of intersection of the line and the ellipse, substitute \( y \) from the line equation into the ellipse equation.

From the line equation:
\[ y = 12 - 3x/4 \]

Substituting into the ellipse equation:
\[ \frac{x^2}{16} + \frac{(12 - 3x/4)^2}{9} = 1 \]

Expanding and simplifying:
1. Substitute \( y \):
\[ \frac{x^2}{16} + \frac{(12 - \frac{3x}{4})^2}{9} = 1 \]
2. Expand the square:
\[ (12 - \frac{3x}{4})^2 = 144 - 2 \cdot 12 \cdot \frac{3x}{4} + \left(\frac{3x}{4}\right)^2 = 144 - 9x + \frac{9x^2}{16} \]
3. Substitute back into the ellipse equation:
\[ \frac{x^2}{16} + \frac{144 - 9x + \frac{9x^2}{16}}{9} = 1 \]

Combine and simplify to find \( x \).

### Step 4: Calculate the area of triangle \( PAB \)
The area \( A \) of triangle \( PAB \) can be calculated using the formula:
\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} \]
Where the base \( AB \) is the distance between points \( A \) and \( B \), and height \( H \) is the perpendicular distance from point \( P \) to line \( AB \).

### Step 5: Set the area equal to \( 6(\sqrt{2}-1) \)
Set the area equal to the given value:
\[ \frac{1}{2} \times AB \times H = 6(\sqrt{2} - 1) \]
From this, solve for \( H \).

### Step 6: Determine the number of points \( P \)
Using the derived height \( H \) and the equation of the line, find the possible angles \( \theta \) that satisfy the conditions for points \( P \) on the ellipse.

### Step 7: Count the solutions
Count the number of valid angles \( \theta \) that satisfy the conditions to find the number of points \( P \).

### Final Answer
The number of such points \( P \) is \( 3 \).

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