If the two curves x=y^2 and xy=k cut each oth | vedclass

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(A) Given curves are $x=y^2 \dots(i)$ and $xy=k \dots(ii)$.From $(i)$, $x=y^2$. Substituting this into $(ii)$, we get $y^2 \cdot y = k$, so $y^3 = k$,

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(A) Given curves are $x=y^2 \dots(i)$ and $xy=k \dots(ii)$.From $(i)$, $x=y^2$. Substituting this into $(ii)$, we get $y^2 \cdot y = k$, so $y^3 = k$, which means $y = k^{1/3}$ and $x = k^{2/3}$.Differentiating $(i)$ with respect to $x$: $1 = 2y \frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{1}{2y} = m_1$.Differentiating $(ii)$ with respect to $x$: $y + x \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{y}{x} = m_2$.Since the curves intersect at right angles, $m_1 m_2 = -1$.$\left(\frac{1}{2y}ight) \left(-\frac{y}{x}ight) = -1 \Rightarrow \frac{1}{2x} = 1 \Rightarrow x = \frac{1}{2}$.Since $x = y^2$, $y^2 = \frac{1}{2}$, so $y = \pm \frac{1}{\sqrt{2}}$.From $xy=k$, we have $k^2 = x^2 y^2 = x^2 (x) = x^3$.Substituting $x = \frac{1}{2}$, $k^2 = (\frac{1}{2})^3 = \frac{1}{8}$.Thus, $8k^2 = 8 	imes \frac{1}{8} = 1$.

If the two curves $x=y^2$ and $xy=k$ cut each other at right angles, then a possible value of $8k^2$ is

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