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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Show that the line $\Large \frac{x}{a}+\frac{y}{b}\normalsize=1$,touches the curve $y=b.e^{\Large \frac{-x}{a}}$ at the point where the curve intersects the axis of y.

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  • If $ y=f(x),$ then $\bigg(\large\frac{dy}{dx} \bigg)_p$ = slope of the tangent to $ y=f(x)$ at point $p$
Step 1
Given equation of the line is $ \large\frac{x}{a}+\large\frac{y}{b}=1\: \: \: (1)$
Equation of the curve is $ y=be^{-\large\frac{x}{a}}$
Since the curve intersects the $y$ axis the coordinate of $x =0$
$ \Rightarrow y=be^{\circ}$
$ \therefore$ The points are $(o, b)$
Step 2
Consider curve $c_1$
differentiating w.r.t $x$ we get,
$ \large\frac{1}{a}+\large\frac{1}{b}\large\frac{dy}{dx}=0$
$ \Rightarrow \large\frac{dy}{dx}=-\large\frac{b}{a}$
Slope of the curve $c_1$ at $(o, b)$
$\bigg( \large\frac{dy}{dx}\bigg)_{(0,b)}=-\large\frac{b}{a}$
Consider the curve $c_2$
differentiating w.r.t $x$ we get,
$ \large\frac{dy}{dx}=-\large\frac{b}{a}e^{-\large\frac{x}{a}}$
The slope of the curve at $(o,b)$ is
$\bigg( \large\frac{dy}{dx}\bigg)_{(o,b)}=-\large\frac{b}{a}e^{\circ}$
$ = -\large\frac{b}{a}$
Hence it is clear that the $ \bigg( \large\frac{dy}{dx}\bigg)_{c_1}=\bigg( \large\frac{dy}{dx}\bigg)_{c_2}$
Hence the curves touch each other at the point $(o,b)$


answered Aug 4, 2013 by thanvigandhi_1

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