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(A) Given curves are $y=10-x^2$ and $y=2+x^2$. To find the point of intersection,equate the two equations: $10-x^2 = 2+x^2 \Rightarrow 2x^2 = 8 \Right

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$\frac{8}{15}$

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(A) Given curves are $y=10-x^2$ and $y=2+x^2$. To find the point of intersection,equate the two equations: $10-x^2 = 2+x^2 \Rightarrow 2x^2 = 8 \Rightarrow x^2 = 4 \Rightarrow x = \pm 2$. For $x=2$,$y=6$. For $x=-2$,$y=6$. Let us consider the point $(2, 6)$. For the first curve $y=10-x^2$,the slope $m_1 = \frac{dy}{dx} = -2x$. At $x=2$,$m_1 = -4$. For the second curve $y=2+x^2$,the slope $m_2 = \frac{dy}{dx} = 2x$. At $x=2$,$m_2 = 4$. The angle $	heta$ between the curves is given by $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$. Substituting the values: $	an 	heta = \left| \frac{-4 - 4}{1 + (-4)(4)} ight| = \left| \frac{-8}{1 - 16} ight| = \left| \frac{-8}{-15} ight| = \frac{8}{15}$. Thus,$|	an 	heta| = \frac{8}{15}$.

If $\theta$ denotes the acute angle between the curves $y=10-x^2$ and $y=2+x^2$,at a point of the intersection,then $|\tan \theta|$ is equal to

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