Show that the straight line `x + y=1` touches the hyperbola `2x ^(2) - 3y ^(2)= 6.` Also find the coordinates of the point of contact.

Answer

To show that the straight line \( x + y = 1 \) touches the hyperbola \( 2x^2 - 3y^2 = 6 \) and to find the coordinates of the point of contact, we can follow these steps:

### Step 1: Rewrite the equations
First, we rewrite the equations of the line and the hyperbola in a more manageable form.

**Line equation:**
\[ y = 1 - x \]

**Hyperbola equation:**
\[ 2x^2 - 3y^2 = 6 \]

### Step 2: Substitute the line equation into the hyperbola equation
Next, we substitute \( y = 1 - x \) into the hyperbola equation.

\[ 2x^2 - 3(1 - x)^2 = 6 \]

### Step 3: Expand and simplify
Now, we expand and simplify the equation.

1. Expand \( (1 - x)^2 \):
\[ (1 - x)^2 = 1 - 2x + x^2 \]

2. Substitute back into the hyperbola equation:
\[ 2x^2 - 3(1 - 2x + x^2) = 6 \]

3. Distribute the -3:
\[ 2x^2 - 3 + 6x - 3x^2 = 6 \]

4. Combine like terms:
\[ -x^2 + 6x - 3 - 6 = 0 \]
\[ -x^2 + 6x - 9 = 0 \]

### Step 4: Rearrange the equation
Rearranging gives:
\[ x^2 - 6x + 9 = 0 \]

### Step 5: Factor the quadratic equation
Now, we can factor the quadratic:
\[ (x - 3)^2 = 0 \]

### Step 6: Solve for \( x \)
Setting the factor equal to zero gives:
\[ x - 3 = 0 \implies x = 3 \]

### Step 7: Find \( y \) using the line equation
Now, we substitute \( x = 3 \) back into the line equation to find \( y \):
\[ y = 1 - 3 = -2 \]

### Step 8: Conclusion
Thus, the coordinates of the point of contact are:
\[ (3, -2) \]

### Step 9: Verify the point lies on the hyperbola
Finally, we can verify that this point lies on the hyperbola:
\[ 2(3)^2 - 3(-2)^2 = 2(9) - 3(4) = 18 - 12 = 6 \]
This confirms that the point \( (3, -2) \) lies on the hyperbola.

### Final Answer
The straight line \( x + y = 1 \) touches the hyperbola \( 2x^2 - 3y^2 = 6 \) at the point \( (3, -2) \).

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