Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

Find the equation of all lines having slope \[0\] which are tangent to the curve \[y = \dfrac{1}{{{x^2} - 2x + 3}}\].

seo-qna
Last updated date: 13th Jun 2024
Total views: 403.5k
Views today: 7.03k
Answer
VerifiedVerified
403.5k+ views
Hint: Using the equation of the curve we will find the slope of the curve at any point at first.
Then equating with the given slope, we will find the point of a tangent.
Using the point of tangent and the slope, we can find the equation of all lines having slope \[0\].

Complete step by step answer:
It is given that; the equation of the given curve is \[y = \dfrac{1}{{{x^2} - 2x + 3}}\].
We have to find the equation of all lines having a slope of \[0\].
Let us consider any point on the curve as \[(x,y)\].
We know that the slope of any curve \[y = f(x)\] at any point is \[f'(x)\].
Here, the curve is \[y = \dfrac{1}{{{x^2} - 2x + 3}}\]
Differentiate with respect to \[x\], we get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - (2x - 2)}}{{{{({x^2} - 2x + 3)}^2}}}\]
Simplifying we get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - 2(x - 1)}}{{{{({x^2} - 2x + 3)}^2}}}\]
It is given that; the slope of the equation is \[0\].
So, according to the problem, we can write,
\[ \Rightarrow \dfrac{{ - 2(x - 1)}}{{{{({x^2} - 2x + 3)}^2}}} = 0\]
Simplifying we get,
\[ \Rightarrow - 2(x - 1) = 0\]
Solving we get,
\[ \Rightarrow x = 1\]
Now, substitute the value of \[x = 1\] in the equation of given curve \[y = \dfrac{1}{{{x^2} - 2x + 3}}\], we will get the value of \[y\].
\[ \Rightarrow y = \dfrac{1}{{{1^2} - 2 \times 1 + 3}} = \dfrac{1}{2}\]
We know that, the equation of a tangent with slope as \[m\] and passing through the point \[({x_1},{y_1})\]is,
\[y - {y_1} = m(x - {x_1})\].
So, the equation of the tangent through \[\left( {1,\dfrac{1}{2}} \right)\] is given by,
\[ \Rightarrow y - \dfrac{1}{2} = 0(x - 1)\]
Solving we get,
\[ \Rightarrow y = \dfrac{1}{2}\]

$\therefore $ The equation of the all lines having slope \[0\] is \[y = \dfrac{1}{2}\].

Note:
The slope (also called Gradient) of a straight line shows how steep a straight line is.
We know that the slope of any curve \[y = f(x)\] at any point is \[f'(x)\].
We know that, the equation of a tangent with slope as \[m\] and passing through the point \[({x_1},{y_1})\] is,
\[y - {y_1} = m(x - {x_1})\].