Consider a parabola `y = (x^(2))/(4)` and the point `F (0,1).`
Let `A _(1)(x_(1), y_(1)),A _(2)(x _(2), y_(2)), A_(3)(x_(3), y_(3)),....., A_(n ) (x _(n), y _(n)) ` are 'n' points on the parabola such `x _(k) gt 0 and angle OFA_(k)= (k pi)/(2n) (k =1,2,3,......, n ).` Then the value of `lim _( n to oo) 1/n sum _(k =1) ^(n) FA_(k),` is equal to :

Options

A. `2/pi`
B. `4/pi`
C. `8/pi`
D. None of these

Answer

To solve the problem, we need to find the limit:

\[ \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} FA_k \]

where \( F(0, 1) \) is a fixed point and \( A_k \) are points on the parabola \( y = \frac{x^2}{4} \) such that the angle \( \angle OFA_k = \frac{k \pi}{2n} \).

### Step 1: Understanding the Parabola
The equation of the parabola can be rewritten as:

\[ x^2 = 4y \]

From this, we can identify that the parameter \( a \) in the standard form \( x^2 = 4ay \) is \( a = 1 \).

### Step 2: Finding Coordinates of Points \( A_k \)
Given that the angle \( \angle OFA_k = \frac{k \pi}{2n} \), we can express the coordinates of the points \( A_k \) in terms of \( k \):

Let \( \theta_k = \frac{k \pi}{2n} \).

Using trigonometric relationships, we can find the coordinates of \( A_k \):

\[ x_k = 2 \tan(\theta_k) \quad \text{and} \quad y_k = \frac{x_k^2}{4} = \tan^2(\theta_k) \]

### Step 3: Finding the Distance \( FA_k \)
The distance \( FA_k \) from point \( F(0, 1) \) to point \( A_k(x_k, y_k) \) is given by:

\[ FA_k = \sqrt{(x_k - 0)^2 + (y_k - 1)^2} = \sqrt{x_k^2 + (y_k - 1)^2} \]

Substituting \( y_k = \tan^2(\theta_k) \):

\[ FA_k = \sqrt{x_k^2 + (\tan^2(\theta_k) - 1)^2} \]

### Step 4: Simplifying \( FA_k \)
Using the identity \( \tan^2(\theta) - 1 = \sec^2(\theta) - 2 \):

\[ FA_k = \sqrt{x_k^2 + (\sec^2(\theta_k) - 2)^2} \]

Substituting \( x_k = 2 \tan(\theta_k) \):

\[ FA_k = \sqrt{(2 \tan(\theta_k))^2 + (\sec^2(\theta_k) - 2)^2} \]

### Step 5: Evaluating the Limit
Now we need to evaluate:

\[ \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} FA_k \]

We can express the sum as an integral in the limit:

\[ \frac{1}{n} \sum_{k=1}^{n} FA_k \approx \int_0^1 FA_k \, dr \]

where \( r = \frac{k}{n} \).

### Step 6: Change of Variables
Let \( t = \frac{\pi r}{4} \), then \( dr = \frac{4}{\pi} dt \) and the limits change from \( 0 \) to \( \frac{\pi}{4} \).

### Step 7: Final Integration
We can now compute the integral:

\[ \int_0^{\frac{\pi}{4}} \sec^2(t) \, dt = \left[ \tan(t) \right]_0^{\frac{\pi}{4}} = \tan\left(\frac{\pi}{4}\right) - \tan(0) = 1 - 0 = 1 \]

### Step 8: Putting it all together
Thus, the final result is:

\[ \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} FA_k = \frac{4}{\pi} \cdot 1 = \frac{4}{\pi} \]

### Conclusion
The value of the limit is:

\[ \frac{4}{\pi} \]