If A and B are foci of ellipse `(x-2y+3)^(2)+(8x +4y +4)^(2) =20` andP is any point on it, then `PA +PB =`

Options

A. 2
B. 4
C. `sqrt(2)`
D. `2sqrt(2)`

Explanation

To solve the problem, we need to find the value of \( PA + PB \) for the ellipse given by the equation:

\[ (x - 2y + 3)^2 + (8x + 4y + 4)^2 = 20 \]

### Step 1: Rewrite the equation in standard form

We start with the given equation:

\[ (x - 2y + 3)^2 + (8x + 4y + 4)^2 = 20 \]

To convert this into the standard form of an ellipse, we divide the entire equation by 20:

\[ \frac{(x - 2y + 3)^2}{20} + \frac{(8x + 4y + 4)^2}{20} = 1 \]

### Step 2: Simplify the terms

Now we can simplify the terms in the equation. Let's rewrite the second term:

\[ \frac{(8x + 4y + 4)^2}{20} = \frac{4(2x + y + 1)^2}{20} = \frac{(2x + y + 1)^2}{5} \]

Now we can rewrite the equation as:

\[ \frac{(x - 2y + 3)^2}{20} + \frac{(2x + y + 1)^2}{5} = 1 \]

### Step 3: Identify \( a^2 \) and \( b^2 \)

From the standard form of the ellipse:

\[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \]

we can identify \( a^2 \) and \( b^2 \) from our equation.

Comparing:

1. \( a^2 = 20 \)
2. \( b^2 = 5 \)

### Step 4: Calculate \( a \)

Now, we find \( a \):

\[ a = \sqrt{20} = 2\sqrt{5} \]

### Step 5: Use the property of the ellipse

In an ellipse, the sum of the distances from any point \( P \) on the ellipse to the two foci \( A \) and \( B \) is equal to \( 2a \):

\[ PA + PB = 2a \]

Substituting the value of \( a \):

\[ PA + PB = 2 \times 2\sqrt{5} = 4\sqrt{5} \]

### Step 6: Conclusion

Thus, the value of \( PA + PB \) is:

\[ \boxed{4\sqrt{5}} \]