Question

Finding Tangents and Normals (a) Find an equation for each tangent to the curve \(y=1 /(x-1)\) that has slope \(-1 .\) (See Exercise \(21 )\) (b) Find an equation for each normal to the curve \(y=1 /(x-1)\)that has slope 1

Short answer

In part (a), the equation of the tangent to the curve \(y = 1/(x-1)\) that has slope -1 is \(y = -x + 3\). In part (b), the equation of the normal to the curve \(y = 1/(x-1)\) that has slope 1 is \(y = x - 1\).

Step-by-step solution

Find the derivative of the curve

The first step is to find the slope of the tangent line to the curve for any point on the curve. This is done by taking the derivative of \(y\). So, finding the derivative of \(y = 1/(x-1)\) using the chain rule, we get \(y' = -1/(x-1)^2\).

Set the derivative equal to the target slope and solve

In part (a), we find those \(x\)-values for which the slope of tangent is -1 (i.e., \(y' = -1\)). After setting \(y' = -1\) and solving, we find \(x = 2\). Hence there is one point on the curve where the slope of the tangent is -1, and that point is \(P(2, 1)\). For part (b), we find those \(x\)-values for which the slope of the normal is 1 (i.e., \(y' = -1\)), getting \(x = 0\). Hence, there is one point on the curve where the slope of the normal is 1, and that point is \(P(0, -1)\).

Write the equations of the lines

The equation of a line with slope \(m\) passing through the point \((x_0, y_0)\) is given by \(y - y_0 = m (x - x_0)\). Using this, in part (a), the equation of the tangent line with slope -1 passing through \(P(2, 1)\) is \(y -1 = -1 (x - 2)\) or \(y = -x + 3\). In part (b), the equation of the normal line with slope 1 passing through \(P(0, -1)\) is \(y + 1 = 1 (x - 0)\) or \(y = x - 1\).