Show that the line `y =x + sqrt7` touches the hyperbola `9x ^(2) - 16 y ^(2) = 144.`

Answer

To show that the line \( y = x + \sqrt{7} \) touches the hyperbola \( 9x^2 - 16y^2 = 144 \), we will follow these steps:

### Step 1: Rewrite the hyperbola in standard form
The given hyperbola equation is:
\[ 9x^2 - 16y^2 = 144 \]
Dividing the entire equation by 144 gives:
\[ \frac{x^2}{16} - \frac{y^2}{9} = 1 \]
This is the standard form of the hyperbola, where \( a^2 = 16 \) and \( b^2 = 9 \).

### Step 2: Identify the slope of the tangent line
The line given is \( y = x + \sqrt{7} \). The slope \( m \) of this line is 1.

### Step 3: Use the formula for the tangent to the hyperbola
The equation of the tangent to the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) at a point where the slope is \( m \) is given by:
\[ y = mx \pm \sqrt{a^2m^2 - b^2} \]
Substituting \( a^2 = 16 \), \( b^2 = 9 \), and \( m = 1 \):
\[ y = x \pm \sqrt{16 \cdot 1^2 - 9} \]
Calculating the square root:
\[ y = x \pm \sqrt{16 - 9} = x \pm \sqrt{7} \]

### Step 4: Compare with the given line
The tangent line can be expressed as:
\[ y = x + \sqrt{7} \quad \text{or} \quad y = x - \sqrt{7} \]
Since the given line is \( y = x + \sqrt{7} \), it matches one of the forms of the tangent line.

### Conclusion
Since the line \( y = x + \sqrt{7} \) can be expressed as a tangent to the hyperbola, we conclude that it touches the hyperbola.