The tangent drawn at any point P to the parab | vedclass

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(D) Let the point $P$ on the parabola ${y^2} = 4ax$ be $(at^2, 2at)$ and the focus $S$ be $(a, 0)$.The equation of the tangent at $P$ is $ty = x + at^

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$90$

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(D) Let the point $P$ on the parabola ${y^2} = 4ax$ be $(at^2, 2at)$ and the focus $S$ be $(a, 0)$.The equation of the tangent at $P$ is $ty = x + at^2$.The directrix of the parabola is $x = -a$.To find the point $K$,substitute $x = -a$ into the tangent equation:$ty = -a + at^2$$y = \frac{a(t^2 - 1)}{t}$So,$K = (-a, \frac{a(t^2 - 1)}{t})$.Now,calculate the slopes of $SP$ and $SK$:Slope of $SP$ $(m_1)$ = $\frac{2at - 0}{at^2 - a} = \frac{2at}{a(t^2 - 1)} = \frac{2t}{t^2 - 1}$.Slope of $SK$ $(m_2)$ = $\frac{\frac{a(t^2 - 1)}{t} - 0}{-a - a} = \frac{a(t^2 - 1)}{t(-2a)} = -\frac{t^2 - 1}{2t}$.Since $m_1 	imes m_2 = (\frac{2t}{t^2 - 1}) 	imes (-\frac{t^2 - 1}{2t}) = -1$,the lines $SP$ and $SK$ are perpendicular.Therefore,the angle $\angle PSK = 90^\circ$.

The tangent drawn at any point $P$ to the parabola ${y^2} = 4ax$ meets the directrix at the point $K$. Then the angle which $KP$ subtends at its focus is ............. $^\circ$.

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