Question

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

Answer 1

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.