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(C) Solving the equations $x^2 + y^2 = 5$ and $y^2 = 4x$ simultaneously:Substituting $y^2 = 4x$ into the circle equation gives $x^2 + 4x - 5 = 0$.Fact

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$3$

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(C) Solving the equations $x^2 + y^2 = 5$ and $y^2 = 4x$ simultaneously:Substituting $y^2 = 4x$ into the circle equation gives $x^2 + 4x - 5 = 0$.Factoring the quadratic,we get $(x + 5)(x - 1) = 0$,so $x = 1$ or $x = -5$.Since $y^2 = 4x$,$x$ must be non-negative,so $x = 1$.For $x = 1$,$y^2 = 4$,which gives $y = \pm 2$. The points of intersection are $(1, 2)$ and $(1, -2)$.For the circle $x^2 + y^2 = 5$,differentiating gives $2x + 2y \frac{dy}{dx} = 0$,so $\frac{dy}{dx} = -\frac{x}{y}$. At $(1, 2)$,$m_1 = -\frac{1}{2}$.For the parabola $y^2 = 4x$,differentiating gives $2y \frac{dy}{dx} = 4$,so $\frac{dy}{dx} = \frac{2}{y}$. At $(1, 2)$,$m_2 = \frac{2}{2} = 1$.The angle $	heta$ between the curves is given by $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$.Substituting the values: $	an 	heta = \left| \frac{-1/2 - 1}{1 + (-1/2)(1)} ight| = \left| \frac{-3/2}{1/2} ight| = |-3| = 3$.

If $\theta$ is the acute angle of intersection at a real point of intersection of the circle $x^2 + y^2 = 5$ and the parabola $y^2 = 4x$,then $\tan \theta$ is equal to

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