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 \( 5x^2 - 9y^2 = 45 \).

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

### Step 2: Expand the equation
Now, we expand \( (x + 2)^2 \):

\[ (x + 2)^2 = x^2 + 4x + 4 \]

Substituting this back into the hyperbola equation gives:

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

### Step 3: Simplify the equation
Now, we simplify the equation:

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

Combine like terms:

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

This simplifies to:

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

### Step 4: Multiply through by -1
To make the equation easier to work with, we multiply through by -1:

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

### Step 5: Factor the quadratic equation
Next, we will factor the quadratic equation. We can rewrite it as:

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

### Step 6: Solve for \( x \)
Setting the factor equal to zero gives:

\[ 2x + 9 = 0 \implies x = -\frac{9}{2} \]

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

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

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

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

### Final Answer
The line \( y = x + 2 \) touches the hyperbola \( 5x^2 - 9y^2 = 45 \) at the point \( \left(-\frac{9}{2}, -\frac{5}{2}\right) \).

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