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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Find the equation of all lines having slope \(2\) that are tangents to the curve $ y= \large\frac{1}{x-3}, $$\: x \neq 3$

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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 : $y=\large\frac{1}{x-3}$
Differentiating w.r.t $x$ we get,
$\large\frac{dy}{dx}=\large\frac{-1}{(x-3)^2}$----(1)
It is given the slope of the tangent is $2$.
Hence equating $\large\frac{dy}{dx}$ to $2$ we get,
$\large\frac{-1}{(x-3)^2}=$$2$
$\Rightarrow -1=2(x-1)^2$
Step 2:
On expanding we get,
$2x^2-12x+18=-1$
$\Rightarrow 2x^2-12+19=0$
Finding a solution for $x$ by using the quadratic formula $\large\frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$\large\frac{12\pm\sqrt{144-4\times 2\times 19}}{4}=\large\frac{12\pm\sqrt{144-158}}{4}$
It has no possible solution.
So the curve has no tangent.
answered Jul 11, 2013 by sreemathi.v
 

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