The tangent and normal at P(t), for all real | vedclass

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(C) The coordinates of $P$ are $(at^2, 2at)$. The tangent at $P$ is $ty = x + at^2$,so its slope $m_1 = \frac{1}{t}$.The normal at $P$ is $y = -tx + 2

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$	an^{-1}t$

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(C) The coordinates of $P$ are $(at^2, 2at)$. The tangent at $P$ is $ty = x + at^2$,so its slope $m_1 = \frac{1}{t}$.The normal at $P$ is $y = -tx + 2at + at^3$. Setting $y=0$,we get $G(a(t^2+2), 0)$.The tangent at $P$ meets the axis at $T(-at^2, 0)$.The circle passes through $P(at^2, 2at)$,$T(-at^2, 0)$,and $G(a(t^2+2), 0)$.The center of the circle is the midpoint of $TG$,which is $(a, 0)$,and the radius is $R = \sqrt{(at^2-a)^2 + (2at)^2} = a(t^2+1)$.The slope of the radius $SP$ is $m_{SP} = \frac{2at-0}{at^2-a} = \frac{2t}{t^2-1}$.The tangent to the circle at $P$ is perpendicular to the radius $SP$,so its slope $m_2 = -\frac{1}{m_{SP}} = \frac{1-t^2}{2t}$.The angle $	heta$ between the tangents is given by $	an 	heta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} ight|$.$	an 	heta = \left| \frac{\frac{1}{t} - \frac{1-t^2}{2t}}{1 + \frac{1}{t} \cdot \frac{1-t^2}{2t}} ight| = \left| \frac{\frac{2-1+t^2}{2t}}{\frac{2t^2+1-t^2}{2t^2}} ight| = \left| \frac{1+t^2}{2t} \cdot \frac{2t^2}{1+t^2} ight| = t$.Therefore,$	heta = 	an^{-1}t$.

The tangent and normal at $P(t),$ for all real positive $t,$ to the parabola $y^2 = 4ax$ meet the axis of the parabola in $T$ and $G$ respectively. Find the angle at which the tangent at $P$ to the parabola is inclined to the tangent at $P$ to the circle passing through the points $P, T,$ and $G$.

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