Question

Find the evolutes of x2/3 y2/3=a2/3

Answer 1

Correct question:

Find the evolute of $x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}$.

Step-by-step explanation:

Let us consider the parametric equations of the given curve,

$x=a\:cos^{3}\theta$

$y=a\:sin^{3}\theta$

By differentiation, we get

$y_{1}=\frac{dy}{dx}=-tan\theta$

$y_{2}=\frac{d^{2}y}{dx^{2}}=\frac{1}{3a}\:sec^{4}\theta\:cosec\theta$

If we take $(\bar{x},\bar{y})$ to be the co-ordinates of the centre of curvature, then

$\bar{x}=x-\frac{y_{1}(1+y_{1}^{2})}{y_{2}}$

$\quad=a\:cos^{3}\theta+\frac{3a\:tan\theta(1+tan^{2}\theta)}{sec^{4}\theta\:cosec\theta}$

$\quad=a\:cos^{3}\theta+3a\:sin^{2}\theta\:cos\theta$ ... (1)

and $\bar{y}=y+\frac{1+y_{1}^{2}}{y_{2}}$

$\quad=a\:sin^{3}\theta+3a\:cos^{2}\theta\:sin\theta$ ... (2)

Here we can obtain the evolute by eliminating $\theta$ from (1) and (2).

Now, $(\bar{x}+\bar{y})=a\:(sin\theta+cos\theta)^{3}$

$\Rightarrow (\bar{x}+\bar{y})^{\frac{2}{3}}=a^{\frac{2}{3}}\:(sin\theta+cos\theta)^{2}$

and similarly, $(\bar{x}-\bar{y})^{\frac{2}{3}}=a^{\frac{2}{3}}\:(cos\theta-sin\theta)^{2}$

$\therefore (\bar{x}+\bar{y}^{\frac{2}{3}}+(\bar{x}-\bar{y})^{\frac{2}{3}}=2a^{\frac{2}{3}}$

Therefore, the required evolute (which is the locus of $(\bar{x},\bar{y})$) is

$(x+y)^{\frac{2}{3}}+(x-y)^{\frac{2}{3}}=2a^{\frac{2}{3}}$

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