Let O be the vertex of the parabola y^2 = 4x | vedclass

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(B) Let the coordinates of points $P$ and $Q$ on the parabola $y^2 = 4x$ be $P(t_1^2, 2t_1)$ and $Q(t_2^2, 2t_2)$.The slope of $OP$ is $m_1 = \frac{2t

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$2$

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(B) Let the coordinates of points $P$ and $Q$ on the parabola $y^2 = 4x$ be $P(t_1^2, 2t_1)$ and $Q(t_2^2, 2t_2)$.The slope of $OP$ is $m_1 = \frac{2t_1}{t_1^2} = \frac{2}{t_1}$ and the slope of $OQ$ is $m_2 = \frac{2t_2}{t_2^2} = \frac{2}{t_2}$.Since $OP \perp OQ$, the product of their slopes is $-1$, so $(\frac{2}{t_1})(\frac{2}{t_2}) = -1$, which implies $t_1t_2 = -4$.Let $M(h, k)$ be the midpoint of $PQ$. Then $h = \frac{t_1^2 + t_2^2}{2}$ and $k = \frac{2t_1 + 2t_2}{2} = t_1 + t_2$.We know that $k^2 = (t_1 + t_2)^2 = t_1^2 + t_2^2 + 2t_1t_2$.Substituting the values, $k^2 = 2h + 2(-4) = 2h - 8$.Thus, the locus of the midpoint is $y^2 = 2(x - 4)$.This is a parabola of the form $y^2 = 4a(x - h')$, where $4a = 2$, so $a = 0.5$.The length of the latus rectum is $4a = 2$.

Let $O$ be the vertex of the parabola $y^2 = 4x$ and its chords $OP$ and $OQ$ are perpendicular to each other. If the locus of the mid-point of the line segment $PQ$ is a conic $C$, then the length of its latus rectum is:

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