If the line `2x+sqrt6y=2` touches the hyperbola `x^(2)-2y^(2)=a^(2)`, then `a^(2)` is equal to

Answer

To solve the problem, we need to find the value of \( a^2 \) such that the line \( 2x + \sqrt{6}y = 2 \) is tangent to the hyperbola \( x^2 - 2y^2 = a^2 \).

### Step-by-Step Solution:

1. **Identify the given line and hyperbola**:
- The line is given by \( 2x + \sqrt{6}y = 2 \).
- The hyperbola is given by \( x^2 - 2y^2 = a^2 \).

2. **Rearrange the line equation**:
- We can express \( y \) in terms of \( x \):
\[ \sqrt{6}y = 2 - 2x \implies y = \frac{2 - 2x}{\sqrt{6}}. \]

3. **Substitute \( y \) into the hyperbola equation**:
- Substitute \( y \) into the hyperbola equation:
\[ x^2 - 2\left(\frac{2 - 2x}{\sqrt{6}}\right)^2 = a^2. \]
- Simplifying the second term:
\[ y^2 = \left(\frac{2 - 2x}{\sqrt{6}}\right)^2 = \frac{(2 - 2x)^2}{6} = \frac{4(1 - x)^2}{3}. \]
- Thus, the hyperbola equation becomes:
\[ x^2 - 2 \cdot \frac{4(1 - x)^2}{3} = a^2. \]
- This simplifies to:
\[ x^2 - \frac{8(1 - 2x + x^2)}{3} = a^2. \]

4. **Combine and simplify**:
- Combine the terms:
\[ x^2 - \frac{8 - 16x + 8x^2}{3} = a^2. \]
- Multiply through by 3 to eliminate the fraction:
\[ 3x^2 - (8 - 16x + 8x^2) = 3a^2. \]
- This simplifies to:
\[ 3x^2 - 8 + 16x - 8x^2 = 3a^2 \implies -5x^2 + 16x - 8 = 3a^2. \]

5. **Set up the quadratic equation**:
- Rearranging gives:
\[ 5x^2 - 16x + (3a^2 + 8) = 0. \]

6. **Condition for tangency**:
- For the line to be tangent to the hyperbola, the discriminant of this quadratic must be zero:
\[ b^2 - 4ac = 0. \]
- Here, \( a = 5 \), \( b = -16 \), and \( c = 3a^2 + 8 \):
\[ (-16)^2 - 4 \cdot 5 \cdot (3a^2 + 8) = 0. \]
- This simplifies to:
\[ 256 - 20(3a^2 + 8) = 0. \]
- Expanding gives:
\[ 256 - 60a^2 - 160 = 0 \implies 96 = 60a^2 \implies a^2 = \frac{96}{60} = \frac{16}{10} = \frac{8}{5}. \]

7. **Final answer**:
- Thus, the value of \( a^2 \) is:
\[ a^2 = \frac{8}{5}. \]