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(B) The equation of the normal to the parabola $y^2 = 4ax$ at point $t$ is $y + tx = 2at + at^3$.The slope of this normal is $m_N = -t$. Since the nor

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$	an^{-1}\left( \frac{1}{2} 	an \phi ight)$

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(B) The equation of the normal to the parabola $y^2 = 4ax$ at point $t$ is $y + tx = 2at + at^3$.The slope of this normal is $m_N = -t$. Since the normal makes an angle $\phi$ with the axis,$m_N = 	an \phi$,so $t = -	an \phi$.If the normal cuts the parabola again at point $t_1$,then $t_1 = -t - \frac{2}{t}$.The tangent at $t_1$ has a slope $m_T = \frac{1}{t_1}$.The angle $	heta$ between the normal and the tangent at the point of intersection is given by $	an 	heta = \left| \frac{m_N - m_T}{1 + m_N m_T} ight| = \left| \frac{-t - 1/t_1}{1 + (-t)(1/t_1)} ight| = \left| \frac{-(tt_1 + 1)}{t_1 - t} ight|$.Substituting $t_1 = -t - \frac{2}{t}$,we get $tt_1 = -t^2 - 2$,so $tt_1 + 1 = -t^2 - 1$.Also $t_1 - t = -2t - \frac{2}{t} = -2(t + \frac{1}{t}) = -\frac{2(t^2+1)}{t}$.Thus,$	an 	heta = \left| \frac{-( -t^2 - 1)}{-\frac{2(t^2+1)}{t}} ight| = \left| \frac{t^2+1}{-\frac{2(t^2+1)}{t}} ight| = \left| -\frac{t}{2} ight| = \frac{|t|}{2}$.Since $t = -	an \phi$,$|t| = |	an \phi|$. Therefore,$	an 	heta = \frac{	an \phi}{2}$.Hence,$	heta = 	an^{-1}\left( \frac{	an \phi}{2} ight)$.

List-$I$	List-$II$
$(A)$ $\left[\frac{p}{2}\left(t+\frac{1}{t}\right), \frac{q}{2}\left(t-\frac{1}{t}\right)\right]$	$(I)$ parabola
$(B)$ $(p+q \cos \theta, r+q \sin \theta)$	$(II)$ circle
$(C)$ $(p+\lambda^2, q-\lambda)$	$(III)$ ellipse
	$(IV)$ hyperbola

If a normal to a parabola $y^2 = 4ax$ makes an angle $\phi$ with its axis,then it will cut the curve again at an angle

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