Question

How many tangent lines to the curve \[y = \dfrac{x}{{x + 1}}\] pass through the point \[\left( {1,2} \right)\]?

Answer 1

Hint: We have to find the tangent lines to the curve \[y = \dfrac{x}{{x + 1}}\] which pass through the point \[\left( {1,2} \right)\]. We will first find the derivative of \[y\] w.r.t \[x\]. Then we will write the equation of tangent using the obtained slope and given point. As the line must touch the curve, we will substitute the given \[y\] in the obtained equation of line to find the tangent lines.

Complete step-by-step answer:
The given curve is \[y = \dfrac{x}{{x + 1}} - - - (1)\].
In this question, we have to find the number of tangent lines to the curve \[y = \dfrac{x}{{x + 1}}\] which passes through the point \[\left( {1,2} \right)\].
For this, we need to calculate the slope of the tangent for the given curve. We know that the first order differentiation of the given curve with respect to \[x\] will be the slope of the tangent at any point.
On differentiating \[(1)\] w.r.t \[x\], we get
\[ \Rightarrow \dfrac{d}{{dx}}\left( y \right) = \dfrac{d}{{dx}}\left( {\dfrac{x}{{x + 1}}} \right)\]
As we know from quotient rule of differentiation, \[\dfrac{d}{{dx}}\left( {\dfrac{u}{v}} \right) = \dfrac{{v\dfrac{{du}}{{dx}} - u\dfrac{{dv}}{{dx}}}}{{{v^2}}}\].
Using this, we get
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\left( {x + 1} \right)\dfrac{d}{{dx}}\left( x \right) - x\dfrac{d}{{dx}}\left( {x + 1} \right)}}{{{{\left( {x + 1} \right)}^2}}}\]
On simplification, we get
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{x + 1 - x}}{{{{\left( {x + 1} \right)}^2}}}\]
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{1}{{{{\left( {x + 1} \right)}^2}}}\]
Therefore, we get slope of the tangent, \[m = \dfrac{1}{{{{\left( {x + 1} \right)}^2}}}\].
According to slope point form of a line, we can write the equation of line passing through \[\left( {1,2} \right)\] and having slope \[m\] is given by,
\[ \Rightarrow y - {y_1} = m\left( {x - {x_1}} \right)\]
Putting the value of \[m\], we get
\[ \Rightarrow y - 2 = \dfrac{1}{{{{\left( {x + 1} \right)}^2}}}\left( {x - 1} \right)\]
On simplifying, we get
\[ \Rightarrow y = \dfrac{{x - 1}}{{{{\left( {x + 1} \right)}^2}}} + 2 - - - (2)\]
As the line must touch the curve, we can substitute \[y = \dfrac{x}{{x + 1}}\].
On substituting \[(1)\] in \[\left( 2 \right)\], we get
\[ \Rightarrow \dfrac{x}{{x + 1}} = \dfrac{{x - 1}}{{{{\left( {x + 1} \right)}^2}}} + 2\]
On taking LCM on RHS, we get
\[ \Rightarrow \dfrac{x}{{x + 1}} = \dfrac{{x - 1 + 2{{\left( {x + 1} \right)}^2}}}{{{{\left( {x + 1} \right)}^2}}}\]
On cross multiplication, we get
\[ \Rightarrow \dfrac{x}{{x + 1}} \times {\left( {x + 1} \right)^2} = x - 1 + 2{\left( {x + 1} \right)^2}\]
\[ \Rightarrow x\left( {x + 1} \right) = x - 1 + 2{\left( {x + 1} \right)^2}\]
On multiplication, we get
\[ \Rightarrow {x^2} + x = x - 1 + 2\left( {{x^2} + 2x + 1} \right)\]
\[ \Rightarrow {x^2} + x = x - 1 + 2{x^2} + 4x + 2\]
On simplifying, we get
\[ \Rightarrow {x^2} + 4x + 1 = 0\]
On solving, we get
\[ \Rightarrow x = \dfrac{{ - 4 \pm \sqrt {{{\left( 4 \right)}^2} - 4 \times 1 \times 1} }}{{2 \times 1}}\]
On simplifying, we get
\[ \Rightarrow x = \dfrac{{ - 4 \pm \sqrt {12} }}{{2 \times 1}}\]
\[ \Rightarrow x = \dfrac{{ - 4 \pm 2\sqrt 3 }}{2}\]
On dividing, we get
\[ \Rightarrow x = - 2 \pm \sqrt 3 \]
Therefore, the tangent lines tangent lines to the curve \[y = \dfrac{x}{{x + 1}}\] which passes through the point \[\left( {1,2} \right)\] is \[x = - 2 + \sqrt 3 \] and \[x = - 2 - \sqrt 3 \].

Note: In geometry, the tangent line or simply tangent to a curve at a given point is the straight line that just touches the curve at a point. It is given by the formula \[y - {y_1} = m\left( {x - {x_1}} \right)\], where \[m\] is the slope and \[{x_1}\] and \[{y_1}\] are the points through which the curve passes. When a point passes through a curve or a line, it satisfies the equation of the curve or the line.