Let PQ be a focal chord of the parabola y^2=4 | vedclass

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(B,D) Let the coordinates of $P$ be $(at^2, 2at)$ and $Q$ be $(a/t^2, -2a/t)$ since $PQ$ is a focal chord.The point of intersection $R$ of the tangent

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$(B, D)$

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(B,D) Let the coordinates of $P$ be $(at^2, 2at)$ and $Q$ be $(a/t^2, -2a/t)$ since $PQ$ is a focal chord.The point of intersection $R$ of the tangents at $P$ and $Q$ is $(a(t \cdot (-1/t)), a(t - 1/t)) = (-a, a(t - 1/t))$.Given $R$ lies on $y = 2x + a$,we substitute $x = -a$ and $y = a(t - 1/t)$:$a(t - 1/t) = 2(-a) + a = -a$$t - 1/t = -1$.$1.$ The length of the focal chord $PQ = a(t + 1/t)^2$.Since $(t + 1/t)^2 = (t - 1/t)^2 + 4 = (-1)^2 + 4 = 5$,$PQ = 5a$. Thus,option $(B)$ is correct.$2.$ The angle $	heta$ subtended by the chord $PQ$ at the vertex $(0,0)$ is given by $	an 	heta = \frac{m_1 - m_2}{1 + m_1m_2}$,where $m_1 = \frac{2at}{at^2} = \frac{2}{t}$ and $m_2 = \frac{-2a/t}{a/t^2} = -2t$.$	an 	heta = \frac{2/t - (-2t)}{1 + (2/t)(-2t)} = \frac{2/t + 2t}{1 - 4} = \frac{2(1/t + t)}{-3}$.Since $(t + 1/t)^2 = 5$,$t + 1/t = \sqrt{5}$ (taking positive root as $t$ is positive for the upper branch).$	an 	heta = \frac{2\sqrt{5}}{-3} = -\frac{2}{3}\sqrt{5}$. Thus,option $(D)$ is correct.

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