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 $xy=a^3$: $y^2 \cdot y = a^3 \Rightarrow y^3 = a^3 \Rightarrow y = a$.Then $x = a^2$. So the point of intersection is $(a^2, a)$.Let $m_1$ and $m_2$ be the slopes of the tangents at $(a^2, a)$.$m_1 = \left(\frac{1}{2y}\right)_{(a^2, a)} = \frac{1}{2a}$.$m_2 = \left(-\frac{y}{x}\right)_{(a^2, a)} = -\frac{a}{a^2} = -\frac{1}{a}$.Since the curves cut orthogonally,$m_1 m_2 = -1$.$\left(\frac{1}{2a}\right) \left(-\frac{1}{a}\right) = -1$.$-\frac{1}{2a^2} = -1$.$2a^2 = 1 \Rightarrow a^2 = \frac{1}{2}$.

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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