Final answer:
To find the equations of the tangent lines to the curve parallel to a given line, we first find the derivative of the curve. Then, we set the derivative equal to the slope of the given line and solve for x. Substituting these values back into the original equation gives us the equations of the tangent lines.
Step-by-step explanation:
First, we need to find the derivative of the given curve. Taking the derivative of y = \frac{{x-9}}{{x+9}}, we get:
y' = \frac{{(x+9)(1)-(x-9)(1)}}{{(x+9)^2}} = \frac{{18}}{{(x+9)^2}}
We know that the equation of a line in slope-intercept form is y = mx + b.
Since we want the tangent lines to be parallel to the line x - y = 9, we know that they must have the same slope. The slope of the given line is 1. Therefore, the slope of the tangent lines to the curve must also be 1.
Setting the derivative equal to 1, we have:
\frac{{18}}{{(x+9)^2}} = 1
From here, we can solve for x and find the corresponding y-values to get the equations of the tangent lines that are parallel to the given line.
After solving for x, we get two values: x = -3 and x = 5. Substituting these values back into the original equation y = \frac{{x-9}}{{x+9}}, we find the corresponding y-values. Therefore, the equations of the tangent lines are:
y = \frac{{x-9}}{{x+9}} for x = -3 and y = \frac{{x-9}}{{x+9}} for x = 5.