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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives

Find the equations of the tangent and normal to the given curves at the indicated points: $ (iv) \: y = x^2\; at \;(0, 0)$

1 Answer

Toolbox:
  • Equation of the tangent at $(x_1,y_1)$ where slope is $m$ is given by $y-y_1=m(x-x_1)$
  • Equation of the normal at $(x_1,y_1)$ where slope is $m$ is given by $y-y_1=\large\frac{-1}{m}$$(x-x_1)$
Step 1:
Given : $y=x^2$
Differentiating w.r.t $x$ we get,
$\large\frac{dy}{dx}$$=2x$
Step 2:
The slope of the tangent at $(0,0)$ is $\large\frac{dy}{dx}_{(0,0)}=$$2(0)$
Therefore $\large\frac{dy}{dx}$$=0$
Therefore the slope of the tangent at $(0,0)$ is $0$.
Hence the equation of the tangent is $(y-0)=0(x-0)$
$y=0$
Step 3:
Equation of the normal at $(0,0)$ is $x=0$
answered Jul 11, 2013 by sreemathi.v
 

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