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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Find the equation of the normal at the point \((am^2,am^3)\) for the curve \(ay^2 = x^3.\)

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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}$$(x-x_1)$
Step 1:
Equation of the curve is $ay^2=x^3$
Differentiating w.r.t $x$ we get,
$2ay\large\frac{dy}{dx}$$=3x^2$
$\Rightarrow \large\frac{dy}{dx}=\frac{3x^2}{2ay}$
Step 2:
At the point $(am^2,am^3)$ the slope is given by
$\large\frac{dy}{dx}_{(am^2,am^3)}=\large\frac{3(am^2)^2}{2a(am^3)}$
$\qquad\qquad\;\;\;\;=\large\frac{3a^2m^4}{2a^2m^3}$
$\qquad\qquad\;\;\;\;=\large\frac{3m}{2}$
Therefore slope of the tangent at $(am^2,am^3)=\large\frac{3m}{2}$
Slope of the normal at $(am^2,am^3)=\large\frac{-2}{3m}$
Step 3:
Hence equation of the normal is $y-am^3=\large\frac{-2}{3m}$$(x-am^2)$
$\Rightarrow 3my-3am^4=-2x+2am^2$
$\Rightarrow 2x+3my-am^2(2+3m^2)=0$
answered Jul 11, 2013 by sreemathi.v
 

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