If the tangents PQ and PR are drawn from a variable point P to the ellipse `(x^(2))/(16)+(y^(2))/(9)=1,` such that the fourth point S of the parallelogram PQSR lies on the circumcircle of the `DeltaPQR`, then the area (in sq. units) of the locus of P is

Options

A. `7pi`
B. `16pi`
C. `25pi`
D. `9pi`

Answer

To solve the problem, we need to find the area of the locus of point P such that the tangents drawn from P to the ellipse \(\frac{x^2}{16} + \frac{y^2}{9} = 1\) create a parallelogram PQSR, where S lies on the circumcircle of triangle PQR.

### Step-by-step Solution:

1. **Understanding the Geometry**:
- The points P, Q, and R are such that PQ and PR are tangents to the ellipse.
- The point S is the fourth vertex of the parallelogram PQSR, which implies that the angle QPR is \(90^\circ\) (since the diagonals of a parallelogram bisect each other).

2. **Setting Up the Coordinates**:
- Let \(P\) be a variable point with coordinates \((h, k)\).
- The equation of the ellipse is given as \(\frac{x^2}{16} + \frac{y^2}{9} = 1\).

3. **Finding the Tangent Equation**:
- The equation of the tangent to the ellipse at point \((h, k)\) can be expressed as:
\[ \frac{hx}{16} + \frac{ky}{9} = 1 \]

4. **Condition for Perpendicular Tangents**:
- Since the angle QPR is \(90^\circ\), the slopes of the tangents from point P must be negative reciprocals of each other.
- Let the slopes of the tangents be \(m_1\) and \(m_2\). Then, \(m_1 \cdot m_2 = -1\).

5. **Using the Tangent Condition**:
- From the tangent condition, we can derive:
\[ h^2 - 16 = k^2 + 9 \]
\[ h^2 - 16 = - (k^2 - 9) \]

6. **Simplifying the Equation**:
- Rearranging gives:
\[ h^2 + k^2 = 25 \]
- This represents a circle centered at the origin with a radius of 5.

7. **Finding the Area of the Locus**:
- The area \(A\) of a circle is given by the formula:
\[ A = \pi r^2 \]
- Here, \(r = 5\), so:
\[ A = \pi \cdot 5^2 = 25\pi \]

### Final Answer:
The area of the locus of point P is \(25\pi\) square units.