Let `f_1 (x,y)-=ax^2+2hxy+by^2=0 ` and let `f_(i+1) (x,y)=0` denote the equation of the bisectors of `f_i (x,y)=0 ` for all i=1,2,3,…..
The value of `sum_(n=2)^(5) (f_(n+2)(x,y))/(f_n(x,y)) is `

Options

A. 14
B. 4
C. 54
D. 6

Answer

To solve the problem, we will follow a systematic approach to find the value of the sum \( \sum_{n=2}^{5} \frac{f_{n+2}(x,y)}{f_n(x,y)} \).

### Step 1: Define the functions
We start with the function:
\[ f_1(x,y) = ax^2 + 2hxy + by^2 = 0 \]

### Step 2: Find the equation of the bisector \( f_2(x,y) \)
The bisector of the conic given by \( f_1(x,y) = 0 \) can be derived using the formula for the angle bisector:
\[ \frac{x^2 - y^2}{a - b} = \frac{xy}{h} \]
Cross-multiplying gives:
\[ h x^2 - y^2 = (a - b) xy \]
Rearranging leads to:
\[ h x^2 - (a - b) xy - y^2 = 0 \]
Thus, we have:
\[ f_2(x,y) = h x^2 - (a - b) xy - y^2 = 0 \]

### Step 3: Find the equation of the bisector \( f_3(x,y) \)
Now, we find \( f_3(x,y) \), which is the bisector of \( f_2(x,y) \):
Using the same angle bisector formula:
\[ \frac{x^2 - y^2}{h} = \frac{xy}{-(a - b)} \]
Cross-multiplying gives:
\[ -(a - b)x^2 - hy^2 = xy \]
Rearranging leads to:
\[ -(a - b)x^2 - hy^2 - xy = 0 \]
Thus, we have:
\[ f_3(x,y) = -(a - b)x^2 - hy^2 - xy = 0 \]

### Step 4: Find the equation of the bisector \( f_4(x,y) \)
Next, we find \( f_4(x,y) \), which is the bisector of \( f_3(x,y) \):
Using the angle bisector formula again:
\[ \frac{x^2 - y^2}{-(a - b)} = \frac{xy}{-h} \]
Cross-multiplying gives:
\[ -hx^2 + y^2 = (a - b)xy \]
Rearranging leads to:
\[ -hx^2 + (a - b)xy - y^2 = 0 \]
Thus, we have:
\[ f_4(x,y) = -hx^2 + (a - b)xy - y^2 = 0 \]

### Step 5: Find the equation of the bisector \( f_5(x,y) \)
Now, we find \( f_5(x,y) \), which is the bisector of \( f_4(x,y) \):
Using the angle bisector formula:
\[ \frac{x^2 - y^2}{-h} = \frac{xy}{-(a - b)} \]
Cross-multiplying gives:
\[ -(a - b)x^2 - hy^2 = xy \]
Rearranging leads to:
\[ -(a - b)x^2 - hy^2 - xy = 0 \]
Thus, we have:
\[ f_5(x,y) = -(a - b)x^2 - hy^2 - xy = 0 \]

### Step 6: Calculate the sum
Now we need to compute:
\[ \sum_{n=2}^{5} \frac{f_{n+2}(x,y)}{f_n(x,y)} \]
This expands to:
\[ \frac{f_4(x,y)}{f_2(x,y)} + \frac{f_5(x,y)}{f_3(x,y)} \]
From our earlier findings, we see that:
- \( f_4(x,y) = f_2(x,y) \)
- \( f_5(x,y) = f_3(x,y) \)

Thus, we have:
\[ \frac{f_4(x,y)}{f_2(x,y)} = 1 \quad \text{and} \quad \frac{f_5(x,y)}{f_3(x,y)} = 1 \]
Adding these gives:
\[ 1 + 1 = 2 \]

### Final Answer
The value of \( \sum_{n=2}^{5} \frac{f_{n+2}(x,y)}{f_n(x,y)} \) is \( 2 \).