If the curves 2x = y^2 and 2xy = K intersect | vedclass

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(D) Given curves are $2x = y^2$ $(i)$ and $2xy = K$ $(ii)$.Solving $(i)$ and $(ii)$,substitute $2x = y^2$ into $(ii)$:$y^2 \cdot y = K \Rightarrow y^3

Vedclass · Accepted Answer

$8$

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(D) Given curves are $2x = y^2$ $(i)$ and $2xy = K$ $(ii)$.Solving $(i)$ and $(ii)$,substitute $2x = y^2$ into $(ii)$:$y^2 \cdot y = K \Rightarrow y^3 = K \Rightarrow y = K^{1/3}$.Then $2x = (K^{1/3})^2 = K^{2/3} \Rightarrow x = \frac{K^{2/3}}{2}$.The intersection point is $(\frac{K^{2/3}}{2}, K^{1/3})$.Differentiating $(i)$ w.r.t. $x$: $2 = 2y \frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{1}{y}$.Slope $m_1$ at the intersection point is $\frac{1}{K^{1/3}}$.Differentiating $(ii)$ w.r.t. $x$: $2(y + x \frac{dy}{dx}) = 0 \Rightarrow \frac{dy}{dx} = -\frac{y}{x}$.Slope $m_2$ at the intersection point is $-\frac{K^{1/3}}{K^{2/3}/2} = -\frac{2K^{1/3}}{K^{2/3}} = -2K^{-1/3}$.Since the curves intersect perpendicularly,$m_1 \cdot m_2 = -1$.$(\frac{1}{K^{1/3}}) \cdot (-2K^{-1/3}) = -1$.$-2K^{-2/3} = -1 \Rightarrow K^{-2/3} = \frac{1}{2}$.$K^{2/3} = 2 \Rightarrow (K^{2/3})^3 = 2^3 \Rightarrow K^2 = 8$.

If the curves $2x = y^2$ and $2xy = K$ intersect perpendicularly,then the value of $K^2$ is

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