For each positive integer consider the point `P` with abscissa `n` on the curve `y^2-x^2=1.` If `d_n` represents the shortest distance from the point `P` to the line `y=x` then `lim_(n->oo)(nd_n)` has the value equal to:

Options

A. (a) `(1)/(2sqrt(2))`
B. (b) `(1)/(2)`
C. (c) `(1)/(sqrt(2))`
D. (d) `0`

Answer

To solve the problem step by step, we will find the limit of \( n d_n \) as \( n \) approaches infinity, where \( d_n \) is the shortest distance from the point \( P(n, \sqrt{n^2 + 1}) \) on the hyperbola \( y^2 - x^2 = 1 \) to the line \( y = x \).

### Step 1: Identify the point \( P \)
The point \( P \) has coordinates \( (n, y) \) where \( y \) satisfies the hyperbola equation:
\[ y^2 - n^2 = 1 \]
This implies:
\[ y^2 = n^2 + 1 \quad \Rightarrow \quad y = \sqrt{n^2 + 1} \]
Thus, the coordinates of point \( P \) are \( (n, \sqrt{n^2 + 1}) \).

### Step 2: Distance from point \( P \) to the line \( y = x \)
The distance \( d_n \) from a point \( (x_0, y_0) \) to the line \( Ax + By + C = 0 \) is given by:
\[ d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} \]
For the line \( y = x \), we can rewrite it as:
\[ x - y = 0 \quad \Rightarrow \quad A = 1, B = -1, C = 0 \]
Thus, the distance \( d_n \) becomes:
\[ d_n = \frac{|n - \sqrt{n^2 + 1}|}{\sqrt{1^2 + (-1)^2}} = \frac{|n - \sqrt{n^2 + 1}|}{\sqrt{2}} \]

### Step 3: Simplifying the expression for \( d_n \)
We need to simplify \( |n - \sqrt{n^2 + 1}| \):
\[ |n - \sqrt{n^2 + 1}| = \sqrt{n^2 + 1} - n \quad \text{(since \( \sqrt{n^2 + 1} > n \))} \]
Now we can substitute this back into our expression for \( d_n \):
\[ d_n = \frac{\sqrt{n^2 + 1} - n}{\sqrt{2}} \]

### Step 4: Further simplification of \( \sqrt{n^2 + 1} - n \)
We can simplify \( \sqrt{n^2 + 1} - n \) by multiplying the numerator and denominator by \( \sqrt{n^2 + 1} + n \):
\[ \sqrt{n^2 + 1} - n = \frac{(\sqrt{n^2 + 1} - n)(\sqrt{n^2 + 1} + n)}{\sqrt{n^2 + 1} + n} = \frac{1}{\sqrt{n^2 + 1} + n} \]
Thus, we have:
\[ d_n = \frac{1}{\sqrt{2}(\sqrt{n^2 + 1} + n)} \]

### Step 5: Finding \( n d_n \)
Now we calculate \( n d_n \):
\[ n d_n = n \cdot \frac{1}{\sqrt{2}(\sqrt{n^2 + 1} + n)} = \frac{n}{\sqrt{2}(\sqrt{n^2 + 1} + n)} \]

### Step 6: Taking the limit as \( n \to \infty \)
We need to evaluate:
\[ \lim_{n \to \infty} n d_n = \lim_{n \to \infty} \frac{n}{\sqrt{2}(\sqrt{n^2 + 1} + n)} \]
As \( n \) approaches infinity, we can approximate \( \sqrt{n^2 + 1} \) as \( n \):
\[ \sqrt{n^2 + 1} \approx n \quad \Rightarrow \quad \sqrt{n^2 + 1} + n \approx 2n \]
Thus:
\[ n d_n \approx \frac{n}{\sqrt{2}(2n)} = \frac{1}{2\sqrt{2}} \]
Therefore, we find:
\[ \lim_{n \to \infty} n d_n = \frac{1}{2\sqrt{2}} \]

### Final Answer
The limit is:
\[ \lim_{n \to \infty} n d_n = \frac{1}{2\sqrt{2}} \]