From a point on the line `y=x+c`, c(parameter), tangents are drawn to the hyperbola `(x^(2))/(2)-(y^(2))/(1)=1` such that chords of contact pass through a fixed point `(x_1, y_1)`. Then , `(x_1)/(y_1)` is equal to

Options

A. `2`
B. `3`
C. `4`
D. None of these

Answer

To solve the problem step by step, we will follow the reasoning provided in the video transcript and derive the solution systematically.

### Step 1: Define the point on the line
Let the point on the line \( y = x + c \) be represented as \( P(h, h + c) \), where \( h \) is a variable.

### Step 2: Write the equation of the hyperbola
The hyperbola is given by the equation:
\[ \frac{x^2}{2} - \frac{y^2}{1} = 1 \]
From this, we can identify \( a^2 = 2 \) and \( b^2 = 1 \).

### Step 3: Write the equation of the chord of contact
The chord of contact from point \( (x_1, y_1) \) to the hyperbola is given by:
\[ \frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1 \]
Substituting \( a^2 \) and \( b^2 \):
\[ \frac{xx_1}{2} - yy_1 = 1 \]

### Step 4: Substitute the coordinates of point \( P \)
For the point \( P(h, h + c) \), we substitute \( x_1 = h \) and \( y_1 = h + c \):
\[ \frac{xh}{2} - y(h + c) = 1 \]

### Step 5: Rearranging the equation
Rearranging gives:
\[ \frac{xh}{2} - yh - yc = 1 \]
This can be rewritten as:
\[ h \left(\frac{x}{2} - y\right) - yc - 1 = 0 \]

### Step 6: Identify the family of lines
This equation represents a family of lines in terms of \( h \). For this family of lines to pass through a fixed point \( (x_1, y_1) \), the coefficient of \( h \) must equal zero:
\[ \frac{x}{2} - y = 0 \quad \text{(1)} \]
and
\[ yc + 1 = 0 \quad \text{(2)} \]

### Step 7: Solve for \( y \) from equation (1)
From equation (1):
\[ \frac{x}{2} = y \quad \Rightarrow \quad y = \frac{x}{2} \]

### Step 8: Substitute \( y \) into equation (2)
Substituting \( y \) into equation (2):
\[ \left(\frac{x}{2}\right)c + 1 = 0 \]
This simplifies to:
\[ \frac{xc}{2} + 1 = 0 \quad \Rightarrow \quad xc = -2 \]

### Step 9: Find the ratio \( \frac{x_1}{y_1} \)
Now, we have \( y_1 = \frac{x_1}{2} \). Thus, substituting into the equation \( xc = -2 \):
\[ x_1c = -2 \quad \Rightarrow \quad c = -\frac{2}{x_1} \]
Substituting \( y_1 = \frac{x_1}{2} \):
\[ x_1 = -2c \quad \Rightarrow \quad \frac{x_1}{y_1} = \frac{x_1}{\frac{x_1}{2}} = 2 \]

### Final Result
Thus, the value of \( \frac{x_1}{y_1} \) is:
\[ \frac{x_1}{y_1} = 2 \]