If `x^(2)+y^(2)=a^(2) and (x^(2))/(16)+(y^(2))/(9)=1` intersect at 4 points P, Q, R and S which form a square, then the area (in sq. units) of the square is

Options

A. `(144)/(25)`
B. `(25)/(4)`
C. `(15)/(2)`
D. `(576)/(25)`

Answer

To solve the problem, we need to find the area of the square formed by the intersection points of the given circle and ellipse. Let's break down the solution step by step.

### Step 1: Identify the equations
We have two equations:
1. Circle: \( x^2 + y^2 = a^2 \)
2. Ellipse: \( \frac{x^2}{16} + \frac{y^2}{9} = 1 \)

### Step 2: Parameterize the points on the circle
The points on the circle can be parameterized as:
\[ (x, y) = (a \cos \theta, a \sin \theta) \]

### Step 3: Substitute the parameterization into the ellipse equation
Substituting \( x = a \cos \theta \) and \( y = a \sin \theta \) into the ellipse equation:
\[ \frac{(a \cos \theta)^2}{16} + \frac{(a \sin \theta)^2}{9} = 1 \]
This simplifies to:
\[ \frac{a^2 \cos^2 \theta}{16} + \frac{a^2 \sin^2 \theta}{9} = 1 \]

### Step 4: Factor out \( a^2 \)
Factoring out \( a^2 \):
\[ a^2 \left( \frac{\cos^2 \theta}{16} + \frac{\sin^2 \theta}{9} \right) = 1 \]
Thus,
\[ a^2 = \frac{1}{\frac{\cos^2 \theta}{16} + \frac{\sin^2 \theta}{9}} \]

### Step 5: Find a common denominator
The common denominator for the fractions is \( 144 \):
\[ \frac{9 \cos^2 \theta + 16 \sin^2 \theta}{144} \]
So,
\[ a^2 = \frac{144}{9 \cos^2 \theta + 16 \sin^2 \theta} \]

### Step 6: Set up the equation for \( a \)
Now we need to find the maximum value of \( a^2 \). To do this, we can analyze the expression \( 9 \cos^2 \theta + 16 \sin^2 \theta \).

### Step 7: Use the method of Lagrange multipliers or analyze critical points
To find the minimum of \( 9 \cos^2 \theta + 16 \sin^2 \theta \), we can differentiate or use trigonometric identities. The minimum occurs when:
\[ \sin^2 \theta = \frac{9}{25} \quad \text{and} \quad \cos^2 \theta = \frac{16}{25} \]
Thus:
\[ 9 \cdot \frac{16}{25} + 16 \cdot \frac{9}{25} = \frac{144}{25} \]

### Step 8: Calculate \( a^2 \)
Substituting back:
\[ a^2 = \frac{144}{\frac{144}{25}} = 25 \]
So, \( a = 5 \).

### Step 9: Find the side length of the square
The side length of the square formed by points P, Q, R, and S can be found by considering the distance from the center of the square to any vertex:
\[ \text{Side length} = 2 \cdot a \cdot \sin(45^\circ) = 2 \cdot 5 \cdot \frac{1}{\sqrt{2}} = \frac{10}{\sqrt{2}} = 5\sqrt{2} \]

### Step 10: Calculate the area of the square
The area \( A \) of the square is given by:
\[ A = (\text{side length})^2 = (5\sqrt{2})^2 = 25 \cdot 2 = 50 \]

### Final Answer
The area of the square formed by points P, Q, R, and S is \( \boxed{50} \) square units.