Question

Find equations of the tangent line to the curve

\(y = \frac{{x - {\bf{1}}}}{{x + {\bf{1}}}}\)

that are parallel to the line \(x - {\bf{2}}y = {\bf{2}}\).

Short answer

The equation of tangents parallel to the line \(x - 2y = 2\) is \(2y = x - 1\) and \(2y = x + 7\).

Step-by-step solution

Find the derivative of the curve using the Quotient rule

The formula for the Quotient rule is \(\left( {\frac{f}{g}} \right)' = \frac{{gf' - fg'}}{{{{\left( g \right)}^2}}}\).

Apply Quotient rule for the function \(y = \frac{{x - 1}}{{x + 1}}\).

\(\begin{aligned}y'\left( x \right) & = \frac{{\rm{d}}}{{{\rm{d}}x}}\left( {\frac{{x - 1}}{{x + 1}}} \right)\\ & = \frac{{\left( {x + 1} \right)\frac{{\rm{d}}}{{{\rm{d}}x}}\left( {x - 1} \right) - \left( {x - 1} \right)\frac{{\rm{d}}}{{{\rm{d}}x}}\left( {x + 1} \right)}}{{{{\left( {x + 1} \right)}^2}}}\\ & = \frac{{\left( {x + 1} \right) - \left( {x - 1} \right)}}{{{{\left( {x + 1} \right)}^2}}}\\ & = \frac{2}{{{{\left( {x + 1} \right)}^2}}}\end{aligned}\)

So, the expression \(\frac{2}{{{{\left( {x + 1} \right)}^2}}}\) represents the slope of the tangent to the curve \(y\left( x \right)\).

Find the coordinate of the point at which tangent is parallel to \(x - {\bf{2}}y = {\bf{2}}\)

The equation of the line is \(x - 2y = 2\), which can also be represented as \(y = \frac{1}{2}x - 1\). By comparing this equation with slope-intercept form \(y = mx + c\), so the slope of the line \(x - 2y = 2\) is \(\frac{1}{2}\).

Let at \(x = a\) the tangent to the given curve is parallel to the line \(x - 2y = 2\). Then, the slope of the tangent is:

\(\begin{aligned}\frac{2}{{{{\left( {a + 1} \right)}^2}}} & = \frac{1}{2}\\{\left( {a + 1} \right)^2} & = 4\\a + 1 & = \pm 2\\a = 1,\, - 3\end{aligned}\)

By the equation of a curve \(y = \frac{{x - 1}}{{x + 1}}\),

For \(a = 1\), \(y = 0\)

For \(a = - 3\), \(y = 2\)

So, the point of tangents are \(\left( {1,0} \right)\) and \(\left( { - 3,2} \right)\).

Find the equation of tangents that are parallel to \(x - {\bf{2}}y = {\bf{2}}\)

The equation of the tangent at \(\left( {1,0} \right)\) with slope \(\frac{1}{2}\) is:

\(\begin{aligned}y - 0 & = \frac{1}{2}\left( {x - 1} \right)\\2y & = x - 1\end{aligned}\)

The equation of the tangent at \(\left( { - 3,2} \right)\) with slope \(\frac{1}{2}\) is:

\(\begin{aligned}{c}y - 2 & = \frac{1}{2}\left( {x + 3} \right)\\2y - 4 & = x + 3\\2y & = x + 7\end{aligned}\)

Thus, the equation of tangents parallel to the line \(x - 2y = 2\) are \(2y = x - 1\) and \(2y = x + 7\).