If two curves x=y^2 and xy=a^3 cut each other | vedclass

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(B) Given curves are $x=y^2$ and $xy=a^3$.Substituting $x=y^2$ into $xy=a^3$,we get $(y^2)y = a^3$,which implies $y^3 = a^3$,so $y=a$.Then $x = y^2 =

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$\frac{1}{2}$

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(B) Given curves are $x=y^2$ and $xy=a^3$.Substituting $x=y^2$ into $xy=a^3$,we get $(y^2)y = a^3$,which implies $y^3 = a^3$,so $y=a$.Then $x = y^2 = a^2$. The point of intersection is $(a^2, a)$.For the curve $x=y^2$,differentiating with respect to $x$: $1 = 2y \frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{1}{2y}$.At $(a^2, a)$,the slope $m_1 = \frac{1}{2a}$.For the curve $xy=a^3$,differentiating with respect to $x$: $x \frac{dy}{dx} + y = 0 \Rightarrow \frac{dy}{dx} = -\frac{y}{x}$.At $(a^2, a)$,the slope $m_2 = -\frac{a}{a^2} = -\frac{1}{a}$.Since the curves cut orthogonally,$m_1 	imes m_2 = -1$.$\left(\frac{1}{2a}ight) 	imes \left(-\frac{1}{a}ight) = -1$.$-\frac{1}{2a^2} = -1$.$2a^2 = 1 \Rightarrow a^2 = \frac{1}{2}$.

If two curves $x=y^2$ and $xy=a^3$ cut each other orthogonally at a point,then $a^2$ is equal to

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