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Prove for an asteroid $x ^{2/3} + y ^{2/3} = a^{2/3}$, the line $\theta = \frac{\pi }{6}$ divide the arc in the first quadrant in a ratio 1:3
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Solution:

Given curve: astroid $x^{2 / 3}+y^{2 / 3}=a^{2 / 3}$

The line $\theta=\pi / 6$ cuts the asroid in 1 st quadrant.

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C is the point on the curve which cuts the arc.

Length of astroid in first quadrant:

Put $\quad x=a \cos ^{3} t \quad$ and $\quad y=a \sin ^{3} t$

$d x=-3 a \sin t . \cos ^{2} t d t \quad, d y=3 \operatorname{acos} t \cdot \sin ^{2} t d t$

$S=\int_{0}^{\frac{\pi}{2}} \sqrt{\left(\frac{d x}{d t}\right)^{2}+\left(\frac{d y}{d t}\right)^{2}}=\int_{0}^{\pi / 2} \sqrt{\left(-3 a \sin t \cdot \cos ^{2} t\right)^{2}+\left(3 a \cos t \cdot \sin ^{2} t\right)^{2}} d t$

$=\int_{0}^{\pi / 2} 3 a \cdot$ sin t.cost dt

$=\frac{3}{2} a \int_{0}^{\pi / 2} \sin 2 t d t$

$=\frac{3}{4} a[-\cos 2 t]_{0}^{\pi / 2}$

$\therefore S=\frac{3}{2} a---(1)$

Now the length of the curve ac: Just put $\frac{\pi}{6}$ insted of $\frac{\pi}{2}$ because the curve is only upto given line.

$\begin{aligned} \therefore \mathrm{S}(\mathrm{ac}) &=\int_{0}^{\pi / 6} 3 a \sin t . \operatorname{cost} d t=\frac{3}{4} a[-\cos 2 t]_{0}^{\pi / 6} \\ &=\frac{3}{4} a\left[-\frac{1}{2}+1\right] \\ \mathrm{S}(\mathrm{ac}) &=\frac{3}{8} a---(2) \end{aligned}$

Legnth of remaining part $=\frac{3}{2} a-\frac{3}{8} a=\frac{9}{8} a \quad \ldots \ldots \ldots \ldots \ldots$ (3)

Divide eqn $(3)$ and $(2)$

The line $\frac{\pi}{6}$ cuts the given astroid in the ratio of $1 : 3$

Hence proved.

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