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Hint: Angle between the curves can be calculated by the formula $\tan \theta = \left| {\dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right|$ . Where, ${m_1}$ and ${m_2}$ are the slopes of the first and second curve respectively.

We know that, angle between the curves can be calculated by the formula $\operatorname{Tan} \theta = \left| {\dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right|$ . Where, ${m_1}$ and ${m_2}$ are the slopes of the first and second curve respectively. Now, to calculate the slope at the point of intersection, we need to find $\dfrac{{dy}}{{dx}}$ for each point.

Let’s calculate the intersection point first. $xy = 8 \Rightarrow x = \dfrac{8}{y},{x^2} = 8y \Rightarrow {(\dfrac{8}{y})^2} = 8y \Rightarrow \dfrac{{64}}{{{y^2}}} = 8y \Rightarrow 8 = {y^3} \Rightarrow y = 2$ .

Using this value of y in $xy = 8$ : $2x = 8 \Rightarrow x = 4$. Hence, the point of intersection will be (4,2). Now, we need to find the derivatives at this point. On differentiating ${x^2} = 8y$ we get, $\dfrac{{d{x^2}}}{{dx}} = 8\dfrac{{dy}}{{dx}} \Rightarrow 2x = 8\dfrac{{dy}}{{dx}} \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{x}{4}$ and on substituting the point ${\dfrac{{dy}}{{dx}}_{(4,2)}} = \dfrac{4}{4} = 1$ .

Again, on differentiation the curve $xy = 8$ we get, $\dfrac{{d(xy)}}{{dx}} = \dfrac{{d8}}{{dx}} \Rightarrow x\dfrac{{dy}}{{dx}} + y\dfrac{{dx}}{{dx}} = 0 \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{y}{x}$ and on substituting the intersection point

${\dfrac{{dy}}{{dx}}_{(4,2)}} = - \dfrac{2}{4} = - \dfrac{1}{2}$ . Now, putting these values to the formula $\tan \theta = \left| {\dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right|$ we get, $\tan \theta = \dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}} \Rightarrow \tan \theta = \dfrac{{ - \dfrac{1}{2} - 1}}{{1 - \dfrac{1}{2}}} \Rightarrow \tan \theta = \dfrac{{ - \dfrac{3}{2}}}{{\dfrac{1}{2}}} \Rightarrow \tan \theta = - \dfrac{3}{{{2}}} \times \dfrac{{{2}}}{1} = - 3 \Rightarrow \theta = {\tan ^{ - 1}}( - 3)$

Note: Here for the differentiation of $xy = 8$ , we have used the formula $\dfrac{{d({y_1}{y_2})}}{{dx}} = {y_1}\dfrac{{d({y_2})}}{{dx}} + {y_2}\dfrac{{d({y_1})}}{{dx}}$ and for the curve ${x^2} = 8y$ , we have used $\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}$.

We know that, angle between the curves can be calculated by the formula $\operatorname{Tan} \theta = \left| {\dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right|$ . Where, ${m_1}$ and ${m_2}$ are the slopes of the first and second curve respectively. Now, to calculate the slope at the point of intersection, we need to find $\dfrac{{dy}}{{dx}}$ for each point.

Let’s calculate the intersection point first. $xy = 8 \Rightarrow x = \dfrac{8}{y},{x^2} = 8y \Rightarrow {(\dfrac{8}{y})^2} = 8y \Rightarrow \dfrac{{64}}{{{y^2}}} = 8y \Rightarrow 8 = {y^3} \Rightarrow y = 2$ .

Using this value of y in $xy = 8$ : $2x = 8 \Rightarrow x = 4$. Hence, the point of intersection will be (4,2). Now, we need to find the derivatives at this point. On differentiating ${x^2} = 8y$ we get, $\dfrac{{d{x^2}}}{{dx}} = 8\dfrac{{dy}}{{dx}} \Rightarrow 2x = 8\dfrac{{dy}}{{dx}} \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{x}{4}$ and on substituting the point ${\dfrac{{dy}}{{dx}}_{(4,2)}} = \dfrac{4}{4} = 1$ .

Again, on differentiation the curve $xy = 8$ we get, $\dfrac{{d(xy)}}{{dx}} = \dfrac{{d8}}{{dx}} \Rightarrow x\dfrac{{dy}}{{dx}} + y\dfrac{{dx}}{{dx}} = 0 \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{y}{x}$ and on substituting the intersection point

${\dfrac{{dy}}{{dx}}_{(4,2)}} = - \dfrac{2}{4} = - \dfrac{1}{2}$ . Now, putting these values to the formula $\tan \theta = \left| {\dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}}} \right|$ we get, $\tan \theta = \dfrac{{{m_1} - {m_2}}}{{1 + {m_1}{m_2}}} \Rightarrow \tan \theta = \dfrac{{ - \dfrac{1}{2} - 1}}{{1 - \dfrac{1}{2}}} \Rightarrow \tan \theta = \dfrac{{ - \dfrac{3}{2}}}{{\dfrac{1}{2}}} \Rightarrow \tan \theta = - \dfrac{3}{{{2}}} \times \dfrac{{{2}}}{1} = - 3 \Rightarrow \theta = {\tan ^{ - 1}}( - 3)$

Note: Here for the differentiation of $xy = 8$ , we have used the formula $\dfrac{{d({y_1}{y_2})}}{{dx}} = {y_1}\dfrac{{d({y_2})}}{{dx}} + {y_2}\dfrac{{d({y_1})}}{{dx}}$ and for the curve ${x^2} = 8y$ , we have used $\dfrac{{d{x^n}}}{{dx}} = n{x^{n - 1}}$.

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