If theta is the acute angle between the curve | vedclass

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(B) To find the angle between the curves $y^2 = x$ and $x^2 + y^2 = 2$, we first find their point of intersection.Substituting $y^2 = x$ into $x^2 + y

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$3$

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(B) To find the angle between the curves $y^2 = x$ and $x^2 + y^2 = 2$, we first find their point of intersection.Substituting $y^2 = x$ into $x^2 + y^2 = 2$, we get $x^2 + x - 2 = 0$.$(x + 2)(x - 1) = 0$, which gives $x = 1$ (since $x = -2$ is not possible for $y^2 = x$).For $x = 1$, $y^2 = 1$, so $y = 1$ (considering the first quadrant intersection).Now, differentiate the curves to find the slopes of the tangents at $(1, 1)$:For $y^2 = x$, $2y \frac{dy}{dx} = 1 \Rightarrow \frac{dy}{dx} = \frac{1}{2y}$. At $(1, 1)$, $m_1 = \frac{1}{2}$.For $x^2 + y^2 = 2$, $2x + 2y \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{x}{y}$. At $(1, 1)$, $m_2 = -1$.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{\frac{1}{2} - (-1)}{1 + (\frac{1}{2})(-1)} ight| = \left| \frac{\frac{3}{2}}{\frac{1}{2}} ight| = 3$.

If $\theta$ is the acute angle between the curves $y^2 = x$ and $x^2 + y^2 = 2$, then $\tan \theta =$

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