Question

The equation of the line parallel to $x$ axis and tangent to the curve $y=\dfrac{1}{{{x}^{2}}+2x+5}$ is

1. $y=\dfrac{1}{4}$
2. $y=4$
3. $y=\dfrac{1}{2}$
4. $y=2$

Answer 1

In this type of question we have to use the concept of derivatives. We know that if a line is tangent to any curve then the slope of the line and the slope of the curve are equal. Also we can find the slope of any curve by taking the derivative of the equation of the curve. So that by equating the slopes we will get the value of $x$ which we substitute in the equation of the curve to obtain the required result.

Complete step-by-step solution:
Now we have to find the equation of line which is parallel to $x$ axis and tangent to the curve $y=\dfrac{1}{{{x}^{2}}+2x+5}$.
For this let us consider
$\Rightarrow y=\dfrac{1}{{{x}^{2}}+2x+5}\cdots \cdots \cdots \left( i \right)$
Let us suppose that the equation of line parallel to $x$ axis is given by $y=k$ where $k$ is some constant.
We know that if a line is tangent to any curve then the slope of the line and the slope of the curve are equal. Thus we can write,
$\Rightarrow \text{Slope of the curve }\left( y=\dfrac{1}{{{x}^{2}}+2x+5} \right)\text{= Slope of the line }\left( y=k \right)$
Now, we can find the slope by taking derivative of both sides with respect to $x$
$\Rightarrow \dfrac{d}{dx}\left( \dfrac{1}{{{x}^{2}}+2x+5} \right)=\dfrac{d}{dx}\left( k \right)$
As we know that, $\dfrac{d}{dx}\left( \dfrac{1}{x} \right)=\dfrac{-1}{{{x}^{2}}}$ and the derivative of a constant is always equal to zero.

& \Rightarrow \dfrac{-1}{{{\left( {{x}^{2}}+2x+5 \right)}^{2}}}\dfrac{d}{dx}\left( {{x}^{2}}+2x+5 \right)=0 \\\ & \Rightarrow \dfrac{-\left( 2x+2 \right)}{{{\left( {{x}^{2}}+2x+5 \right)}^{2}}}=0 \\\ & \Rightarrow -2x-2=0 \\\ & \Rightarrow -2x=2 \\\ & \Rightarrow x=-1 \\\ \end{aligned}$$ Now by substituting this value of $$x$$ in $$\left( i \right)$$ and on simplification we get, $$\begin{aligned} & y=\dfrac{1}{{{x}^{2}}+2x+5} \\\ & \Rightarrow y=\dfrac{1}{{{\left( -1 \right)}^{2}}+2\left( -1 \right)+5} \\\ & \Rightarrow y=\dfrac{1}{1-2+5} \\\ & \Rightarrow y=\dfrac{1}{-1+5} \\\ & \therefore y=\dfrac{1}{4} \\\ \end{aligned}$$ Hence the equation of the line parallel to $$x$$ axis and tangent to the curve $$y=\dfrac{1}{{{x}^{2}}+2x+5}$$ is $$y=\dfrac{1}{4}$$ Thus option (1) is the correct option. **Note:** In this type of question students have to note that the first derivative of any curve will give the slope of the tangent line of the curve. Students have to take care in calculation of the first derivative, they have to remember that $$\dfrac{d}{dx}\left( \dfrac{1}{f\left( x \right)} \right)=\dfrac{-1}{{{\left( f\left( x \right) \right)}^{2}}}\dfrac{d}{dx}f\left( x \right)$$. Also students have to remember that the equation of the line parallel to $$x$$ axis is given by $$y=k$$ where $$k$$ is some constant.