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How do you find the equations of the tangent lines to the curve $y = \dfrac{{x - 1}}{{x + 1}}$ that are parallel to the line $x - 2y = 2?$

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Last updated date: 20th Jun 2024
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Answer
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Hint:Here we will use the concepts of the algebra and the arithmetic equations. Also, we will concentrate on the reasoning and the calculus. Remember the slope of the two parallel lines are always equal.

Complete step by step solution:
A line which is parallel to the given line is $x - 2y = 2$ must have the same slope. Comparing the given equation with the standard line equation: $y = mx + c$
Take the given expression: $x - 2y = 2$
Convert the above equation in the form of the standard equation.
$ - 2y = 2 - x$
Taking constant on the opposite side.
$ \Rightarrow y = - 1 + \dfrac{x}{2}$
So, the slope of the given line is equal to $\dfrac{1}{2}$
Now, the slope of the tangent line can be derived using the derivation.
Now, the given expression is $y = \dfrac{{x - 1}}{{x + 1}}$ …(A)
So, here we will apply $\left( {\dfrac{u}{v}} \right)$rule to find the derivation.
$y' = \dfrac{{x(x - 1) - (x - 1).1}}{{{{\left( {x + 1} \right)}^2}}}$
Simplify the above equation, like terms with the same value and the opposite sign cancel each other.
$y' = \dfrac{2}{{{{\left( {x + 1} \right)}^2}}}$
Now, to find the value of “x”, take $y' = \dfrac{1}{2}$ in the above equation.
$\dfrac{1}{2} = \dfrac{2}{{{{\left( {x + 1} \right)}^2}}}$
Take cross multiplication, where the numerator of one side is multiplied with the denominator of the opposite side and vice-versa.
$ \Rightarrow {(x + 1)^2} = 4$
Take the square root on both the sides of the equation.
$ \Rightarrow \sqrt {{{(x + 1)}^2}} = \sqrt 4 $
Square and square root cancel each other on the left hand side of the equation.
$ \Rightarrow x + 1 = \pm 2$
So, $x + 1 = 2$ or $x + 1 = - 2$
When you move any term from one side of the equation to the other side, then the sign of the term changes. Positive terms become negative and vice-versa.
$ \Rightarrow x = 1$ or $x = ( - 3)$
Now, placing the values of “x” in the equation (A)
When $x = 1,$
$ \Rightarrow y = 0$
And when $x = ( - 3)$
$ \Rightarrow y = 2$
Therefore, the equation of the line $(1,0)$ with the slope $m = \dfrac{1}{2}$ and equation of the line through $( - 3,2)$ with the slope $m = \dfrac{1}{2}$ using the $y - {y_0} = m(x - {x_0})$
Which gives –
$y = \dfrac{1}{2}x + \dfrac{7}{2}$ and
$y = \dfrac{1}{2}x - \dfrac{1}{2}$


Note: Be careful about the sign convention and the simplification of the algebraic expressions. Always remember when you move any term from one side to another sign of the term changes. Positive terms become negative and vice-versa.