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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Find the equation of all lines having slope $–1$ that are tangents to the curve $ y= \large\frac{1}{x-1}, $$ \: x \neq 1$

$\begin{array}{1 1} (A)\;x+y+1=0 ; x+y-3=0 \\ (B)\;x+y-1=0 ; x+y-3=0 \\ (C)\;x+y+1=0 ; x-y-3=0 \\ (D)\;x-y-1=0 ; x+y-3=0 \end{array} $

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1 Answer

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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 :$y=\large\frac{1}{x-1}$
Differentiating with respect to $x$ we get,
$\large\frac{dy}{dx}=\frac{-1}{(x-1)^2}$-----(1)
The given slope is $-1$
Therefore $\large\frac{dy}{dx}$$=-1$----(2)
Step 2:
Now equating equ(1) and (2) we get,
$\large\frac{-1}{(x-1)^2}=$$-1$
$\Rightarrow (x-1)^2=1$
$x-1=\pm 1$
$\Rightarrow x=2,0$
Step 3:
When $x=2,y=1$
When $x=0,y=-1$
Equation of a tangent is $y-y_1=m(x-x_1)$
Let $(x_1,y_1)$ be $(2,1)$ and $m=-1$,then
$y-1=-1(x-2)$
$\Rightarrow y-1=-x+2$
$x+y-3=0$
Step 4:
Let $(x_1,y_1)$ be $(0,-1)$ and $m=-1$,then
$y-(-1)=-1(x-0)$
$\Rightarrow y+1=-x$
$x+y+1=0$
Hence the required lines are $x+y+1=0$ and $x+y-3=0$
answered Jul 10, 2013 by sreemathi.v
 

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