A tangent to the hyperbola `y = (x+9)/(x+5)` passing through the origin is

Options

A. `x+ 25y =0`
B. `5x+y =0`
C. `5x-y =0`
D. `x -25y =0`

Explanation

To find the tangent to the hyperbola given by the equation \( y = \frac{x + 9}{x + 5} \) that passes through the origin, we can follow these steps:

### Step 1: Rewrite the Hyperbola Equation
The hyperbola can be rewritten as:
\[ y = \frac{x + 9}{x + 5} = 1 + \frac{4}{x + 5} \]
This makes it easier to analyze the function.

### Step 2: Find the Derivative
To find the slope of the tangent line at any point on the hyperbola, we need to differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{d}{dx} \left( 1 + \frac{4}{x + 5} \right) \]
Using the quotient rule, we get:
\[ \frac{dy}{dx} = 0 - \frac{4 \cdot 1}{(x + 5)^2} = -\frac{4}{(x + 5)^2} \]

### Step 3: Equation of the Tangent Line
The equation of the tangent line at a point \( (x_1, y_1) \) on the hyperbola can be expressed as:
\[ y - y_1 = m(x - x_1) \]
where \( m \) is the slope at that point. Substituting \( y_1 = 1 + \frac{4}{x_1 + 5} \) and \( m = -\frac{4}{(x_1 + 5)^2} \), we have:
\[ y - \left(1 + \frac{4}{x_1 + 5}\right) = -\frac{4}{(x_1 + 5)^2}(x - x_1) \]

### Step 4: Substitute the Origin into the Tangent Equation
Since the tangent passes through the origin (0, 0), we substitute \( x = 0 \) and \( y = 0 \):
\[ 0 - \left(1 + \frac{4}{x_1 + 5}\right) = -\frac{4}{(x_1 + 5)^2}(0 - x_1) \]
This simplifies to:
\[ -1 - \frac{4}{x_1 + 5} = \frac{4x_1}{(x_1 + 5)^2} \]

### Step 5: Clear the Denominator
Multiply through by \( (x_1 + 5)^2 \) to eliminate the fraction:
\[ -(x_1 + 5)^2 - 4(x_1 + 5) = 4x_1 \]
Expanding this gives:
\[ -(x_1^2 + 10x_1 + 25) - 4x_1 - 20 = 4x_1 \]
Combining like terms results in:
\[ -x_1^2 - 18x_1 - 45 = 4x_1 \]
Rearranging gives:
\[ -x_1^2 - 22x_1 - 45 = 0 \]
Multiplying through by -1:
\[ x_1^2 + 22x_1 + 45 = 0 \]

### Step 6: Solve the Quadratic Equation
Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x_1 = \frac{-22 \pm \sqrt{22^2 - 4 \cdot 1 \cdot 45}}{2 \cdot 1} = \frac{-22 \pm \sqrt{484 - 180}}{2} = \frac{-22 \pm \sqrt{304}}{2} \]
Simplifying gives:
\[ x_1 = \frac{-22 \pm 4\sqrt{19}}{2} = -11 \pm 2\sqrt{19} \]

### Step 7: Find Corresponding \( y_1 \) Values
Substituting \( x_1 \) back into the hyperbola equation to find \( y_1 \):
\[ y_1 = 1 + \frac{4}{x_1 + 5} \]

### Step 8: Write the Tangent Equations
For each \( x_1 \), we can find the corresponding \( y_1 \) and write the tangent equations using the point-slope form.

### Final Answer
The tangents to the hyperbola passing through the origin are:
1. \( x + 25y = 0 \)
2. \( x + y = 0 \)