The equation of the tangent parallel to `y-x+5=0" drawan to "(x^(2))/(3)-(y^(2))/(2)=1` is

Options

A. `x-y-1=0`
B. `x-y+2=0`
C. `x+y-1=0`
D. x+y+2=0`

Answer

To find the equation of the tangent parallel to \( y - x + 5 = 0 \) drawn to the hyperbola given by \( \frac{x^2}{3} - \frac{y^2}{2} = 1 \), we can follow these steps:

### Step 1: Identify the slope of the given line
The equation of the line can be rewritten in slope-intercept form:
\[ y = x - 5 \]
From this, we see that the slope \( m \) of the line is \( 1 \).

**Hint:** The slope-intercept form of a line is \( y = mx + c \), where \( m \) is the slope.

### Step 2: Write the equation of the tangent line
Since we need a line parallel to the given line, we can express the equation of the tangent line in the form:
\[ y = x + c \]
where \( c \) is a constant that we need to determine.

**Hint:** Parallel lines have the same slope.

### Step 3: Use the condition for tangency
The equation of the hyperbola is given by:
\[ \frac{x^2}{3} - \frac{y^2}{2} = 1 \]
For the line \( y = x + c \) to be tangent to the hyperbola, we substitute \( y \) in the hyperbola's equation:
\[ \frac{x^2}{3} - \frac{(x + c)^2}{2} = 1 \]

**Hint:** Substitute the expression for \( y \) into the hyperbola's equation.

### Step 4: Expand and simplify
Expanding the equation:
\[ \frac{x^2}{3} - \frac{x^2 + 2cx + c^2}{2} = 1 \]
This simplifies to:
\[ \frac{x^2}{3} - \frac{x^2}{2} - cx - \frac{c^2}{2} = 1 \]
To combine the terms, find a common denominator (which is 6):
\[ \frac{2x^2}{6} - \frac{3x^2}{6} - cx - \frac{c^2}{2} = 1 \]
This leads to:
\[ -\frac{x^2}{6} - cx - \frac{c^2}{2} - 1 = 0 \]
Multiplying through by -6 gives:
\[ x^2 + 6cx + 3c^2 + 12 = 0 \]

**Hint:** When simplifying, ensure all terms are combined correctly.

### Step 5: Apply the condition for tangency
For the line to be tangent to the hyperbola, the discriminant of the quadratic equation must be zero:
\[ (6c)^2 - 4(1)(3c^2 + 12) = 0 \]
Calculating the discriminant:
\[ 36c^2 - 12c^2 - 48 = 0 \]
This simplifies to:
\[ 24c^2 - 48 = 0 \]
Solving for \( c^2 \):
\[ 24c^2 = 48 \implies c^2 = 2 \implies c = \pm \sqrt{2} \]

**Hint:** The discriminant must equal zero for there to be exactly one solution (tangency).

### Step 6: Write the equations of the tangents
Substituting \( c \) back into the tangent line equation:
\[ y = x + \sqrt{2} \quad \text{and} \quad y = x - \sqrt{2} \]
These can be rewritten in standard form:
\[ x - y + \sqrt{2} = 0 \quad \text{and} \quad x - y - \sqrt{2} = 0 \]

**Hint:** Rearranging the equation gives the standard form of the line.

### Final Answer
The equations of the tangents parallel to \( y - x + 5 = 0 \) drawn to the hyperbola are:
\[ x - y + \sqrt{2} = 0 \quad \text{and} \quad x - y - \sqrt{2} = 0 \]