If the area enclosed between the curves y = k | vedclass

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Question

$y=k x^{2}, x=k y^{2}$

$\Rightarrow x=k\left(k^{2} x^{4}\right) $

$\Rightarrow x=0$ or $x^{3}=\left(\frac{1}{k}\right)^{3}$

$ \Rightarrow x=\

Vedclass · Accepted Answer

$\frac{1}{{\sqrt 3 }}$

Vedclass · Answer

$y=k x^{2}, x=k y^{2}$

$\Rightarrow x=k\left(k^{2} x^{4}\right) $

$\Rightarrow x=0$ or $x^{3}=\left(\frac{1}{k}\right)^{3}$

$ \Rightarrow x=\frac{1}{k}, 0$

Point of intersection are

$\left(\frac{1}{\mathrm{k}}, \frac{1}{\mathrm{k}}\right)$ and $(0,0)$

Area $\int\limits_0^{1/k} {\left( {\sqrt {\frac{x}{k}}  - k{x^\prime }} \right)dx}  = 1$

$ \Rightarrow {\left( {\frac{1}{{\sqrt k }}\frac{{{x^{3/2}}}}{{3/2}} - \frac{{k{x^2}}}{3}} \right)^{1/k}}$

$=1 \Rightarrow \frac{2}{3 k^{2}}=1$

$ \Rightarrow k^{2}=\frac{1}{3}$

$k=\frac{1}{\sqrt{3}}$

If the area enclosed between the curves $y = kx^2$ and $x = ky^2, (k > 0)$, is $1$ square unit. Then $k$ is

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