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(B) Given curves are $y = x$ and $y^2 = 4x$. For $y = x$,differentiating with respect to $x$,we get $m_1 = \frac{dy}{dx} = 1$. For $y^2 = 4x$,differen

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$	an^{-1}\left(\frac{1}{3}ight)$

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(B) Given curves are $y = x$ and $y^2 = 4x$. For $y = x$,differentiating with respect to $x$,we get $m_1 = \frac{dy}{dx} = 1$. For $y^2 = 4x$,differentiating with respect to $x$,we get $2y \frac{dy}{dx} = 4$,which implies $m_2 = \frac{dy}{dx} = \frac{2}{y}$. At the point $(4, 4)$,the slope $m_1 = 1$ and $m_2 = \frac{2}{4} = \frac{1}{2}$. The angle of intersection $	heta$ 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{1/2}{3/2} ight| = \frac{1}{3}$. Therefore,$	heta = 	an^{-1}\left(\frac{1}{3}ight)$.

List-$I$	List-$II$
$I$. Vertex	$A$. $8$
$II$. Length of latus rectum	$B$. $(\frac{29}{10}, \frac{-38}{10})$
$III$. Directrix	$C$. $3x+4y-1=0$
$IV$. One end of the latus rectum	$D$. $(\frac{-2}{5}, \frac{-16}{5})$
	$E$. $6$

What is the angle of intersection between the curves $y = x$ and $y^2 = 4x$ at the point $(4, 4)$?

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