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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Find the co-ordinates of the point on the curve $\sqrt x\quad \sqrt y=4$ at which tangent is equally inclined to the axes.

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  • If $y=f(x),$ then $\bigg(\large\frac{dy}{dx}\bigg)_p=$ slope of the tangent to $y=f(r)$ at point $p$
Let the required points be $(x_1 , y_1 )$
Given equation of the curve is
$ \sqrt x + \sqrt y=4$
Since $(x_1, y_1)$ lies on the curve we get,
$\sqrt {x_1}+\sqrt{y_1}=4$
Consider $\sqrt x+\sqrt y=4$
differentiating w.r.t $x$ we get,
$ \large\frac{1}{2\sqrt x}+\large\frac{1}{2\sqrt y}\large\frac{dy}{dx}=0$
$ \Rightarrow \large\frac{dy}{dx}=-\large\frac{\sqrt y}{\sqrt x}$
Step 2
It is given that the tangent is equally inclined to the axes.
$ \Rightarrow \theta=45^{\circ}$
$ \therefore \tan\: \theta=1$
$ \Rightarrow \bigg( \large\frac{dy}{dx} \bigg)_{(x_1,y_1)}=1$
$ \Rightarrow - \large\frac{\sqrt {y_1}}{\sqrt{x_1}}=1$
$ \therefore -\sqrt{y_1}=\sqrt{x_1}$
Squaring on both sides we get,
$x_1=y_1$
$ \therefore \sqrt{x_1}+\sqrt{y_1}=4$
Substituting for $y_1$ we get
$2\sqrt{x_1}=4$
Squaring on both sides we get
$4x_1=16$
$\Rightarrow x_1=4 \: and \: y_1=4$
Hence the required points are $(4, 4)$

 

answered Aug 4, 2013 by thanvigandhi_1
edited Aug 4, 2013 by thanvigandhi_1
 

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