The line `y=x+2` touches the hyperbola `5x^2-9y^2=45` at the point

Options

A. (0,2)
B. (3,1)
C. `(-(9)/(2),-(5)/(2))`
D. None of these

Answer

To find the point at which the line \( y = x + 2 \) touches the hyperbola \( 5x^2 - 9y^2 = 45 \), we will follow these steps:

### Step 1: Substitute the line equation into the hyperbola equation
We start by substituting \( y = x + 2 \) into the hyperbola equation:

\[ 5x^2 - 9y^2 = 45 \]

Substituting \( y \):

\[ 5x^2 - 9(x + 2)^2 = 45 \]

### Step 2: Expand the equation
Now, we expand the equation:

\[ 5x^2 - 9(x^2 + 4x + 4) = 45 \]

This simplifies to:

\[ 5x^2 - 9x^2 - 36x - 36 = 45 \]

### Step 3: Combine like terms
Combining like terms gives us:

\[ -4x^2 - 36x - 36 = 45 \]

Now, move 45 to the left side:

\[ -4x^2 - 36x - 36 - 45 = 0 \]

This simplifies to:

\[ -4x^2 - 36x - 81 = 0 \]

### Step 4: Multiply through by -1
To make calculations easier, we can multiply the entire equation by -1:

\[ 4x^2 + 36x + 81 = 0 \]

### Step 5: Factor the quadratic equation
Next, we will factor the quadratic equation. Notice that it can be written as a perfect square:

\[ (2x + 9)^2 = 0 \]

### Step 6: Solve for x
Setting the expression equal to zero gives:

\[ 2x + 9 = 0 \]

Solving for \( x \):

\[ 2x = -9 \implies x = -\frac{9}{2} \]

### Step 7: Find the corresponding y value
Now, we substitute \( x = -\frac{9}{2} \) back into the line equation to find \( y \):

\[ y = x + 2 = -\frac{9}{2} + 2 \]

Converting 2 to a fraction gives:

\[ y = -\frac{9}{2} + \frac{4}{2} = -\frac{5}{2} \]

### Step 8: Write the point of tangency
Thus, the point at which the line touches the hyperbola is:

\[ \left(-\frac{9}{2}, -\frac{5}{2}\right) \]

### Conclusion
The answer is \( \left(-\frac{9}{2}, -\frac{5}{2}\right) \).

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