If the segment of the parabola y^2 = 4ax inte | vedclass

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Question

(C) Let the point on the parabola be $P(at^2, 2at)$.The tangent at $P$ is $ty = x + at^2$.The directrix is $x = -a$.To find the intersection point $Q$

Vedclass · Accepted Answer

$\frac{\pi}{2}$

Vedclass · Answer

(C) Let the point on the parabola be $P(at^2, 2at)$.The tangent at $P$ is $ty = x + at^2$.The directrix is $x = -a$.To find the intersection point $Q$ of the tangent and the directrix,substitute $x = -a$ into the tangent equation:$ty = -a + at^2 \implies y = \frac{a(t^2 - 1)}{t}$.So,$Q = (-a, \frac{a(t^2 - 1)}{t})$.The focus is $S(a, 0)$.The slope of $SP$ is $m_1 = \frac{2at - 0}{at^2 - a} = \frac{2t}{t^2 - 1}$.The slope of $SQ$ is $m_2 = \frac{\frac{a(t^2 - 1)}{t} - 0}{-a - a} = \frac{a(t^2 - 1)}{t(-2a)} = -\frac{t^2 - 1}{2t}$.Since $m_1 	imes m_2 = \left(\frac{2t}{t^2 - 1}ight) 	imes \left(-\frac{t^2 - 1}{2t}ight) = -1$,the lines $SP$ and $SQ$ are perpendicular.Therefore,$	heta = \frac{\pi}{2}$.

If the segment of the parabola $y^2 = 4ax$ intercepted by its directrix subtends an angle $\theta$ at its focus,then $\theta = \dots$

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