Find the equation of tangent to the hyperbola `y=(x+9)/(x+5) ` which passes through `(0, 0)` origin

Options

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

Explanation

To find the equation of the tangent to the hyperbola given by \( y = \frac{x + 9}{x + 5} \) that passes through the origin (0, 0), we can follow these steps:

### Step 1: Rewrite the Hyperbola Equation
The hyperbola can be rewritten as:
\[ y = \frac{x + 9}{x + 5} \]
This can also be expressed as:
\[ y = 1 + \frac{4}{x + 5} \]

### Step 2: Find the Slope of the Tangent
Let the point of tangency be \( P(x_1, y_1) \). The slope of the tangent to the hyperbola at this point can be found using the derivative:
\[ \frac{dy}{dx} = -\frac{4}{(x + 5)^2} \]
At the point \( P(x_1, y_1) \), the slope becomes:
\[ m = -\frac{4}{(x_1 + 5)^2} \]

### Step 3: Write the Equation of the Tangent Line
The equation of the tangent line at point \( P(x_1, y_1) \) can be expressed as:
\[ y - y_1 = m(x - x_1) \]
Substituting \( y_1 = \frac{x_1 + 9}{x_1 + 5} \) and \( m \):
\[ y - \frac{x_1 + 9}{x_1 + 5} = -\frac{4}{(x_1 + 5)^2}(x - x_1) \]

### Step 4: Substitute the Origin into the Tangent Equation
Since the tangent line passes through the origin (0, 0), we substitute \( x = 0 \) and \( y = 0 \):
\[ 0 - \frac{x_1 + 9}{x_1 + 5} = -\frac{4}{(x_1 + 5)^2}(0 - x_1) \]
This simplifies to:
\[ -\frac{x_1 + 9}{x_1 + 5} = \frac{4x_1}{(x_1 + 5)^2} \]

### Step 5: Clear the Denominators
Multiplying both sides by \( (x_1 + 5)^2 \) gives:
\[ -(x_1 + 9)(x_1 + 5) = 4x_1 \]
Expanding the left side:
\[ -x_1^2 - 14x_1 - 45 = 4x_1 \]
Rearranging the equation:
\[ -x_1^2 - 18x_1 - 45 = 0 \]
Multiplying through by -1:
\[ x_1^2 + 18x_1 + 45 = 0 \]

### Step 6: Solve the Quadratic Equation
Now we can solve the quadratic equation using the quadratic formula:
\[ x_1 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Where \( a = 1, b = 18, c = 45 \):
\[ x_1 = \frac{-18 \pm \sqrt{18^2 - 4 \cdot 1 \cdot 45}}{2 \cdot 1} \]
Calculating the discriminant:
\[ = \frac{-18 \pm \sqrt{324 - 180}}{2} = \frac{-18 \pm \sqrt{144}}{2} = \frac{-18 \pm 12}{2} \]
This gives us two solutions:
\[ x_1 = \frac{-6}{2} = -3 \quad \text{and} \quad x_1 = \frac{-30}{2} = -15 \]

### Step 7: Find the Tangent Equations for Each \( x_1 \)
1. For \( x_1 = -3 \):
- Calculate \( y_1 \):
\[ y_1 = \frac{-3 + 9}{-3 + 5} = \frac{6}{2} = 3 \]
- The slope \( m \):
\[ m = -\frac{4}{(-3 + 5)^2} = -\frac{4}{4} = -1 \]
- The equation of the tangent:
\[ y - 3 = -1(x + 3) \implies y = -x \]

2. For \( x_1 = -15 \):
- Calculate \( y_1 \):
\[ y_1 = \frac{-15 + 9}{-15 + 5} = \frac{-6}{-10} = \frac{3}{5} \]
- The slope \( m \):
\[ m = -\frac{4}{(-15 + 5)^2} = -\frac{4}{100} = -\frac{1}{25} \]
- The equation of the tangent:
\[ y - \frac{3}{5} = -\frac{1}{25}(x + 15) \implies 25y - 15 = -x - 15 \implies x + 25y = 0 \]

### Final Result
The equations of the tangents to the hyperbola that pass through the origin are:
1. \( y = -x \)
2. \( x + 25y = 0 \)