The angle between the curves x^2=8y and xy=8 | vedclass

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(B) Given equations of the curves are $x^2=8y$ $(i)$ and $xy=8$ $(ii)$.To find the point of intersection,substitute $y = \frac{x^2}{8}$ from $(i)$ int

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$	an^{-1}(3)$

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(B) Given equations of the curves are $x^2=8y$ $(i)$ and $xy=8$ $(ii)$.To find the point of intersection,substitute $y = \frac{x^2}{8}$ from $(i)$ into $(ii)$:$x\left(\frac{x^2}{8}ight) = 8 \Rightarrow x^3 = 64 \Rightarrow x = 4$.Substituting $x=4$ into $(ii)$,we get $4y=8 \Rightarrow y=2$.So,the point of intersection is $(4, 2)$.For curve $(i)$,$x^2=8y \Rightarrow 2x = 8 \frac{dy}{dx} \Rightarrow \frac{dy}{dx} = \frac{x}{4}$.At $(4, 2)$,the slope $m_1 = \frac{4}{4} = 1$.For curve $(ii)$,$xy=8 \Rightarrow x \frac{dy}{dx} + y = 0 \Rightarrow \frac{dy}{dx} = -\frac{y}{x}$.At $(4, 2)$,the slope $m_2 = -\frac{2}{4} = -\frac{1}{2}$.The angle $	heta$ between the curves is given by $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$.$	an 	heta = \left| \frac{1 - (-1/2)}{1 + (1)(-1/2)} ight| = \left| \frac{3/2}{1/2} ight| = 3$.Thus,$	heta = 	an^{-1}(3)$. Note: The angle between curves is typically defined as the acute angle,so $	an^{-1}(3)$ is the standard representation.

The angle between the curves $x^2=8y$ and $xy=8$ is

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