logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
0 votes

Find the equation of all lines having slope \(0\) that are tangents to the curve $ y=\large \frac{1}{x^2-2x+3}$

$\begin{array}{1 1} y = 0 \\ y=\frac{-1}{2} \\y=\frac{1}{2} \\ \text{No tangent} \end{array} $

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • If $y=f(x)$,then $\big(\large\frac{dy}{dx}\big)_P$=slope of the tangent to $y=f(x)$ at the point $P$.
  • Equation of line with given two points is $\large\frac{x-x_1}{x_2-x_1}=\large\frac{y-y_1}{y_2-y_1}$
Step 1:
Given : $y=\large\frac{1}{x^2-2x+3}$
Differentiating w.r.t $x$ we get,
$\large\frac{dy}{dx}=\frac{(x^2-2x+3).0-1(2x-2)}{(x^2-2x+3)^2}$
$\large\frac{dy}{dx}=\frac{-(2x-2)}{(x^2-2x+3)^2}$
Step 2:
But the slope is given as 0.
Therefore $\large\frac{-(2x-2)}{(x^2-2x+3)^2}$$=0$
$\Rightarrow -2(x-1)=0$
$\Rightarrow x=1$
Step 3:
When $x=1,y=\large\frac{1}{1^2-2(1)+3}$
$\Rightarrow \large\frac{1}{2}$
Therefore the tangent at $(1,\large\frac{1}{2})$ having slope $0$ is
$(y-y_1)=m(x-x_1)$
$\Rightarrow (y-\large\frac{1}{2})=$$0(x-1)$
$\Rightarrow y-\large\frac{1}{2}$$=0$
$y=\large\frac{1}{2}$ is the required equation of the tangent.
answered Jul 11, 2013 by sreemathi.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...