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The slope of tangent o the curve $x=t^2+3t-8,y=2t^2-2t-5$ at the point (2,-1) is :\[(A)\;\frac{22}{7}\quad(B)\;\frac{6}{7}(C)\;{-6}{7}\quad(D)\;-6\]

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  • Slope of a line is $ ax+by+c=0$ is $ - \bigg( \large\frac{ Coefficient\: of \: x}{ Coefficient\: of \: y} \bigg)$
  • If $ y = f(x),\: then \: \bigg( \large\frac{dy}{dx} \bigg)_p = $ slope of the normal to $ y = f(x)$ at point $p$
Step 1
The parameters of the curve are
$ x = t^2+3t-8$
differentiating w.r.t $t$ we get,
$ \large\frac{dx}{dt}=2t+3$
and $y=2t^2-2t-5$
differentiating w.r.t $t$ we get,
$ \large\frac{dy}{dt}=4t-2$
$ \therefore \large\frac{dy}{dt}=\large\frac{dy}{dt} \times \large\frac{dt}{dx}$
$ = \large\frac{4t-2}{2t+3}$
Step 2
when $x=2$
then $t$ is
$ \Rightarrow t^2+3t-10=0$
$ \Rightarrow (t+5)(t-2)=0$
$ \Rightarrow t=-5\: or \: t=2$
when $t=2$
$ \large\frac{dy}{dx} = \large\frac{4(2)-2}{2(2)+3}$
$= \large\frac{6}{7}$
Hence the correct option is B
answered Aug 11, 2013 by thanvigandhi_1

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