The two curves x=y^2 and xy=a^3 cut orthogona | vedclass

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(B) Given curves are $x=y^2$ $(i)$ and $xy=a^3$ (ii).For curve $(i)$, differentiating with respect to $x$: $1 = 2y \frac{dy}{dx} \Rightarrow \frac{dy}

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$\frac{1}{2}$

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(B) Given curves are $x=y^2$ $(i)$ and $xy=a^3$ (ii).For curve $(i)$, differentiating with respect to $x$: $1 = 2y \frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{1}{2y}$.For curve (ii), differentiating with respect to $x$: $x \frac{dy}{dx} + y = 0 \Rightarrow \frac{dy}{dx} = -\frac{y}{x}$.To find the point of intersection, substitute $x=y^2$ into $xy=a^3$: $y^2(y) = a^3 \Rightarrow y^3 = a^3 \Rightarrow y = a$.Then $x = a^2$. So the point of intersection is $(a^2, a)$.The slopes at $(a^2, a)$ are $m_1 = \frac{1}{2a}$ and $m_2 = -\frac{a}{a^2} = -\frac{1}{a}$.Since the curves cut orthogonally, $m_1 m_2 = -1$.Therefore, $(\frac{1}{2a})(-\frac{1}{a}) = -1 \Rightarrow -\frac{1}{2a^2} = -1 \Rightarrow 2a^2 = 1 \Rightarrow a^2 = \frac{1}{2}$.

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

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