Tangent is drawn at the point `(x_(i), y_(i))` on the curve y = f(x), which intersects the x-axis at `(x_(i+1), 0)`. Now again tangent is drawn at `(x_(i+1), y_(i+1))` on the curve which intersects the x-axis at `(x_(i+2), 0)` and the process is repeated n times i.e. i = 1, 2, 3,......,n.
If `x_(1), x_(2), x_(3),...,x_(n)` form a geometric progression with common ratio equal to 2 and the curve passes through (1, 2), then the curve is

Options

A. circle
B. hyperbola
C. ellipse
D. parabola

Explanation

To solve the problem, we need to find the curve \( y = f(x) \) that satisfies the given conditions. Let's break down the steps:

### Step 1: Understand the Geometric Progression
We are given that \( x_1, x_2, x_3, \ldots, x_n \) form a geometric progression with a common ratio of 2. This means:
\[ x_2 = 2x_1, \quad x_3 = 2x_2 = 4x_1, \quad x_4 = 2x_3 = 8x_1, \ldots \]
In general, we can express this as:
\[ x_i = 2^{i-1} x_1 \]

### Step 2: Find the Tangent Line Equation
At each point \( (x_i, y_i) \), the tangent line can be expressed using the derivative \( f'(x_i) \):
\[ y - y_i = f'(x_i)(x - x_i) \]
Setting \( y = 0 \) (to find where the tangent intersects the x-axis), we get:
\[ 0 - f(x_i) = f'(x_i)(x_{i+1} - x_i) \]
This simplifies to:
\[ x_{i+1} = x_i - \frac{f(x_i)}{f'(x_i)} \]

### Step 3: Substitute the Geometric Progression
Substituting \( x_i = 2^{i-1} x_1 \) into the tangent line equation:
\[ x_{i+1} = 2^{i-1} x_1 - \frac{f(2^{i-1} x_1)}{f'(2^{i-1} x_1)} \]
Since \( x_{i+1} = 2^i x_1 \), we have:
\[ 2^i x_1 = 2^{i-1} x_1 - \frac{f(2^{i-1} x_1)}{f'(2^{i-1} x_1)} \]
Rearranging gives:
\[ \frac{f(2^{i-1} x_1)}{f'(2^{i-1} x_1)} = -x_1 \]

### Step 4: Analyze the Function
Let \( k = x_1 \). The equation becomes:
\[ f(2^{i-1} k) = -k f'(2^{i-1} k) \]

### Step 5: Assume a Form for \( f(x) \)
Assume \( f(x) = ax^b \) for some constants \( a \) and \( b \). Then:
\[ f'(x) = abx^{b-1} \]
Substituting into the equation:
\[ f(2^{i-1} k) = a(2^{i-1} k)^b = a \cdot 2^{(i-1)b} k^b \]
\[ f'(2^{i-1} k) = ab(2^{i-1} k)^{b-1} = ab \cdot 2^{(i-1)(b-1)} k^{b-1} \]
Thus, we have:
\[ a \cdot 2^{(i-1)b} k^b = -k \cdot ab \cdot 2^{(i-1)(b-1)} k^{b-1} \]
Simplifying gives:
\[ a \cdot 2^{(i-1)b} = -ab \cdot 2^{(i-1)(b-1)} \]
This leads to:
\[ a \cdot 2^{(i-1)b} + ab \cdot 2^{(i-1)(b-1)} = 0 \]
Factoring out \( 2^{(i-1)(b-1)} \):
\[ 2^{(i-1)(b-1)}(a \cdot 2^{(i-1)} + ab) = 0 \]
This implies:
\[ a \cdot 2^{(i-1)} + ab = 0 \Rightarrow a(2^{(i-1)} + b) = 0 \]
Since \( a \neq 0 \), we have:
\[ b = -2^{(i-1)} \]

### Step 6: Determine the Function
Since the curve passes through \( (1, 2) \), we can find \( a \):
\[ f(1) = a \cdot 1^b = 2 \Rightarrow a = 2 \]
Thus, the function is:
\[ f(x) = 2x^{-2} \]

### Final Answer
The curve is:
\[ y = \frac{2}{x^2} \]