If theta is the angle between the curves y^2= | vedclass

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(C) First, find the points of intersection of the curves $y^2=4x$ and $x^2+y^2=5$. Substituting $y^2=4x$ into the second equation gives $x^2+4x-5=0$.

Vedclass · Accepted Answer

$3$

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(C) First, find the points of intersection of the curves $y^2=4x$ and $x^2+y^2=5$. Substituting $y^2=4x$ into the second equation gives $x^2+4x-5=0$. Factoring the quadratic equation, 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 $x=1$, $y^2=4(1)=4$, which gives $y=2$ or $y=-2$. We consider the point $(1, 2)$. For the curve $y^2=4x$, differentiating with respect to $x$ gives $2y \frac{dy}{dx} = 4$, so $\frac{dy}{dx} = \frac{2}{y}$. At $(1, 2)$, $m_1 = \frac{2}{2} = 1$. For the curve $x^2+y^2=5$, differentiating with respect to $x$ gives $2x + 2y \frac{dy}{dx} = 0$, so $\frac{dy}{dx} = -\frac{x}{y}$. At $(1, 2)$, $m_2 = -\frac{1}{2}$. The angle $	heta$ between the curves is given by $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$. Substituting the values, $	an 	heta = \left| \frac{1 - (-1/2)}{1 + (1)(-1/2)} ight| = \left| \frac{3/2}{1/2} ight| = 3$. Thus, $|	an 	heta| = 3$.

If $\theta$ is the angle between the curves $y^2=4x$ and $x^2+y^2=5$, then $|\tan \theta|=$

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