`2x+sqrt(6)y=2` touches the hyperbola `x^(2)-2y^(2)=4`, then the point of contact is

Options

A. `((1)/(2),(1)/(sqrt(6)))`
B. `(4,-sqrt(6))`
C. `(4,sqrt(6))`
D. `(-2,sqrt(6))`

Answer

To find the point of contact of the line \( 2x + \sqrt{6}y = 2 \) with the hyperbola \( x^2 - 2y^2 = 4 \), we can follow these steps:

### Step 1: Write the equations
We have the line equation:
\[ 2x + \sqrt{6}y = 2 \]
And the hyperbola equation:
\[ x^2 - 2y^2 = 4 \]

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

### Step 3: Substitute \( y \) into the hyperbola equation
Now, substitute \( y \) into the hyperbola equation:
\[ x^2 - 2\left(\frac{2 - 2x}{\sqrt{6}}\right)^2 = 4 \]

### Step 4: Simplify the equation
Calculating \( \left(\frac{2 - 2x}{\sqrt{6}}\right)^2 \):
\[ \left(\frac{2 - 2x}{\sqrt{6}}\right)^2 = \frac{(2 - 2x)^2}{6} = \frac{4(1 - x)^2}{6} = \frac{2(1 - x)^2}{3} \]
Substituting this back into the hyperbola equation:
\[ x^2 - 2 \cdot \frac{2(1 - x)^2}{3} = 4 \]
\[ x^2 - \frac{4(1 - 2x + x^2)}{3} = 4 \]
Multiplying through by 3 to eliminate the fraction:
\[ 3x^2 - 4(1 - 2x + x^2) = 12 \]
Expanding:
\[ 3x^2 - 4 + 8x - 4x^2 = 12 \]
Combining like terms:
\[ -x^2 + 8x - 16 = 0 \]
Multiplying through by -1:
\[ x^2 - 8x + 16 = 0 \]

### Step 5: Factor the quadratic equation
This factors to:
\[ (x - 4)^2 = 0 \]
Thus, we have:
\[ x = 4 \]

### Step 6: Find \( y \) using the value of \( x \)
Substituting \( x = 4 \) back into the equation for \( y \):
\[ y = \frac{2 - 2(4)}{\sqrt{6}} = \frac{2 - 8}{\sqrt{6}} = \frac{-6}{\sqrt{6}} = -\sqrt{6} \]

### Step 7: State the point of contact
The point of contact is:
\[ (4, -\sqrt{6}) \]

### Final Answer
The point of contact is \( (4, -\sqrt{6}) \).

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