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Find equation of normal to the curve $ay^2 = x^3$ at the point whose $x$ coordinate is am².

Answers (1)

Since $x$ coordinate is $\text{am}^2$ , so $ay^2 = ({am}^2)^3 = a^2m^6$

$\Rightarrow y = \pm am^3$

$\therefore$ point is $P(am^2, am^3)$ and $Q(am^2, -am^3)$

Now differentiating the given curve w.r.t $x$ we get:

$2ay\frac{dy}{dx} = 3x^2 \Rightarrow \frac{dy}{dx} = \frac{3x^2}{2ay}$

$\Rightarrow\left . \frac{dy}{dx}\right ]_{\text {at P}} = \frac{3a^2m^4}{2a^2m^3} = \frac{3m}{2}$ and $\Rightarrow\left . \frac{dy}{dx}\right ]_{\text {at Q}} = \frac{3a^2m^4}{-2a^2m^3} = \frac{-3m}{2}$

$\therefore m_{N} \text{ at P} = \frac{-2}{3m}$ and $m_{N} \text{ at Q} = \frac{2}{3m}$

Equation of normal at P: $y - am^3 = \frac{-2}{3m}(x-am^2)$

$\Rightarrow 2x + 3my -3am^4 - 2am^2 = 0$

Equation of normal at Q: $y + am^3 = \frac{2}{3m}(x-am^2)$

$\Rightarrow 2x - 3my -3am^4 - 2am^2 = 0$

Posted by

Ravindra Pindel

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