`A , r = 1 , 2, 3 ….., n ` are n points on the parabola `y^(2)=4x` in the first quadrant . If `A_(r) = (x_(r),y_(r))` where `x_(1),x_(2),….x_(n)` are in G.P and `x_(1)=1,x_(2)=2 ` then `y_(n)` is equal to

Options

A. `2^((n+1)/2)`
B. `2^(n+1)`
C. `(sqrt(2))^(n+1)`
D. `2^(n//2)`

Answer

To solve the problem step by step, we will follow the given information about the points on the parabola \( y^2 = 4x \) and the geometric progression (G.P.) of the x-coordinates.

### Step 1: Identify the first two terms of the G.P.
Given:
- \( x_1 = 1 \)
- \( x_2 = 2 \)

Since \( x_1 \) and \( x_2 \) are the first two terms of a G.P., we can find the common ratio \( r \):
\[ r = \frac{x_2}{x_1} = \frac{2}{1} = 2 \]

### Step 2: Find the general term of the G.P.
The general term \( x_n \) of a G.P. can be expressed as:
\[ x_n = x_1 \cdot r^{n-1} \]
Substituting the values of \( x_1 \) and \( r \):
\[ x_n = 1 \cdot 2^{n-1} = 2^{n-1} \]

### Step 3: Use the parabola equation to find \( y_n \)
The points \( (x_r, y_r) \) lie on the parabola defined by the equation \( y^2 = 4x \). For \( x_n \):
\[ y_n^2 = 4x_n \]
Substituting \( x_n = 2^{n-1} \):
\[ y_n^2 = 4 \cdot 2^{n-1} \]
This simplifies to:
\[ y_n^2 = 2^2 \cdot 2^{n-1} = 2^{n+1} \]

### Step 4: Solve for \( y_n \)
Taking the square root of both sides:
\[ y_n = \sqrt{2^{n+1}} = 2^{(n+1)/2} \]

### Step 5: Final expression for \( y_n \)
Thus, we can express \( y_n \) as:
\[ y_n = 2^{(n+1)/2} \]

### Conclusion
The value of \( y_n \) is \( 2^{(n+1)/2} \).

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