A chord is drawn passing through `P(2,2)` on the ellipse `(x^(2))/(25)+(y^(2))/(16) =1` such that it intersects the ellipse at A and B. Then maximum value of `PA.PB` is

Options

A. `(61)/(4)`
B. `(59)/(4)`
C. `(71)/(4)`
D. `(63)/(4)`

Explanation

To solve the problem, we need to find the maximum value of the product of distances \( PA \) and \( PB \) where \( A \) and \( B \) are the points where the chord intersects the ellipse and \( P(2, 2) \) is a point on the chord.

### Step-by-step Solution:

1. **Equation of the Ellipse**:
The equation of the ellipse is given by:
\[ \frac{x^2}{25} + \frac{y^2}{16} = 1 \]
Here, \( a^2 = 25 \) and \( b^2 = 16 \), so \( a = 5 \) and \( b = 4 \).

2. **Parametric Form of the Ellipse**:
The parametric equations for the ellipse can be expressed as:
\[ x = 5 \cos \theta, \quad y = 4 \sin \theta \]

3. **Equation of the Chord**:
The chord passing through point \( P(2, 2) \) can be expressed in parametric form:
\[ \frac{x - 2}{\cos \theta} = \frac{y - 2}{\sin \theta} = r \]
From this, we can express \( x \) and \( y \) in terms of \( r \):
\[ x = r \cos \theta + 2, \quad y = r \sin \theta + 2 \]

4. **Substituting into the Ellipse Equation**:
Substitute \( x \) and \( y \) into the ellipse equation:
\[ \frac{(r \cos \theta + 2)^2}{25} + \frac{(r \sin \theta + 2)^2}{16} = 1 \]

5. **Expanding and Rearranging**:
Expanding the equation gives:
\[ \frac{r^2 \cos^2 \theta + 4r \cos \theta + 4}{25} + \frac{r^2 \sin^2 \theta + 4r \sin \theta + 4}{16} = 1 \]
Multiply through by 400 (LCM of 25 and 16) to eliminate the denominators:
\[ 16(r^2 \cos^2 \theta + 4r \cos \theta + 4) + 25(r^2 \sin^2 \theta + 4r \sin \theta + 4) = 400 \]

6. **Combining Terms**:
This simplifies to:
\[ (16 \cos^2 \theta + 25 \sin^2 \theta) r^2 + (64 \cos \theta + 100 \sin \theta) r + (64 + 100 - 400) = 0 \]
Which can be rearranged to:
\[ (16 \cos^2 \theta + 25 \sin^2 \theta) r^2 + (64 \cos \theta + 100 \sin \theta) r - 236 = 0 \]

7. **Using the Quadratic Formula**:
The product of the roots \( r_1 \) and \( r_2 \) (which correspond to points A and B) is given by:
\[ r_1 r_2 = \frac{-c}{a} = \frac{236}{16 \cos^2 \theta + 25 \sin^2 \theta} \]

8. **Finding Maximum Value of \( PA \cdot PB \)**:
To maximize \( PA \cdot PB \), we need to minimize the denominator:
\[ 16 \cos^2 \theta + 25 \sin^2 \theta \]
Let \( t = \sin^2 \theta \), then \( \cos^2 \theta = 1 - t \):
\[ 16(1 - t) + 25t = 16 + 9t \]
The minimum value occurs when \( t = 0 \) (i.e., \( \theta = 0 \)):
\[ \text{Minimum value} = 16 \]

9. **Final Calculation**:
Thus, the maximum value of \( PA \cdot PB \) is:
\[ PA \cdot PB = \frac{236}{16} = \frac{59}{4} \]

### Conclusion:
The maximum value of \( PA \cdot PB \) is:
\[ \frac{59}{4} \]