Find the equations of both lines that are tangent to the curve 𝑦𝑦 = 𝑥𝑥−1
𝑥𝑥+1 and are parallel to the line 𝑥𝑥 − 2𝑦𝑦 = 2.
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We are tasked with finding the equations of the two lines that are tangent to the curve \( y = \frac{x - 1}{x + 1} \) and are parallel to the line \( x - 2y = 2 \).

### Step 1: Determine the slope of the given line

The line \( x - 2y = 2 \) can be rewritten in slope-intercept form \( y = mx + b \). Solving for \( y \):

\[
x - 2y = 2 \implies -2y = -x + 2 \implies y = \frac{1}{2}x - 1
\]

Thus, the slope of this line is \( m = \frac{1}{2} \).

### Step 2: Find the derivative of the curve to get the slope of the tangent lines

The derivative of the curve \( y = \frac{x - 1}{x + 1} \) gives us the slope of the tangent line at any point on the curve. We use the quotient rule to differentiate:

\[
y = \frac{x - 1}{x + 1}
\]
Let \( u = x - 1 \) and \( v = x + 1 \), so:
\[
\frac{dy}{dx} = \frac{v \cdot \frac{du}{dx} - u \cdot \frac{dv}{dx}}{v^2}
\]
where \( \frac{du}{dx} = 1 \) and \( \frac{dv}{dx} = 1 \). Substituting these into the quotient rule:

\[
\frac{dy}{dx} = \frac{(x + 1)(1) - (x - 1)(1)}{(x + 1)^2} = \frac{x + 1 - x + 1}{(x + 1)^2} = \frac{2}{(x + 1)^2}
\]

So, the slope of the tangent line at any point on the curve is \( \frac{2}{(x + 1)^2} \).

### Step 3: Set the slope of the tangent line equal to the slope of the given line

We want the slope of the tangent line to equal the slope of the given line, which is \( \frac{1}{2} \). Thus, we set:

\[
\frac{2}{(x + 1)^2} = \frac{1}{2}
\]

### Step 4: Solve for \( x \)

Multiply both sides by \( (x + 1)^2 \) and simplify:

\[
2 = \frac{1}{2}(x + 1)^2 \implies (x + 1)^2 = 4
\]

Taking the square root of both sides:

\[
x + 1 = \pm 2
\]

So, we have two solutions:

\[
x + 1 = 2 \implies x = 1
\]
\[
x + 1 = -2 \implies x = -3
\]

### Step 5: Find the corresponding \( y \)-coordinates

Now, we substitute these \( x \)-values back into the original curve equation to find the corresponding \( y \)-values:

- For \( x = 1 \):

\[
y = \frac{1 - 1}{1 + 1} = 0
\]

- For \( x = -3 \):

\[
y = \frac{-3 - 1}{-3 + 1} = \frac{-4}{-2} = 2
\]

### Step 6: Write the equations of the tangent lines

Now that we have the points of tangency \( (1, 0) \) and \( (-3, 2) \), and the slope is \( \frac{1}{2} \), we can use the point-slope form to write the equations of the tangent lines:

- For the point \( (1, 0) \):

\[
y - 0 = \frac{1}{2}(x - 1) \implies y = \frac{1}{2}x - \frac{1}{2}
\]

- For the point \( (-3, 2) \):

\[
y - 2 = \frac{1}{2}(x + 3) \implies y = \frac{1}{2}x + \frac{7}{2}
\]

### Final Answer

The equations of the two tangent lines are:

\[
y = \frac{1}{2}x - \frac{1}{2}
\]
and
\[
y = \frac{1}{2}x + \frac{7}{2}
\]

Would you like more details or have any questions about this solution? Here are five related questions:

1. How is the quotient rule applied in differentiation?
2. Can you explain the point-slope form of a line?
3. How do we interpret the slope of a line geometrically?
4. What are other ways to find tangent lines to curves?
5. Can a curve have more than two tangent lines parallel to a given line?

**Tip:** The quotient rule is useful whenever you are differentiating a ratio of two functions, like in this problem!

Find the equations of both lines that are tangent to the curve y = (x - 1) / (x + 1) and are parallel to the line x - 2y = 2.

Discover how to find the equations of two tangent lines to the curve y = (x - 1) / (x + 1) that are parallel to the line x - 2y = 2. This solution involves calculus concepts like differentiation using the quotient rule and identifying the slopes of tangent lines. Perfect for students in Grades 11-12.

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