Let P and Q be points on the parabola y^{2}=4 | vedclass

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(C) Let the coordinates of points $P$ and $Q$ on the parabola $y^{2}=4x$ be $(t^{2}, 2t)$ and $(m^{2}, 2m)$ respectively. The vertex of the parabola i

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$4$

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(C) Let the coordinates of points $P$ and $Q$ on the parabola $y^{2}=4x$ be $(t^{2}, 2t)$ and $(m^{2}, 2m)$ respectively. The vertex of the parabola is $X(0, 0)$.Since $PQ$ subtends a right angle at the vertex $X$, the product of the slopes of $XP$ and $XQ$ is $-1$.Slope of $XP = \frac{2t-0}{t^{2}-0} = \frac{2}{t}$.Slope of $XQ = \frac{2m-0}{m^{2}-0} = \frac{2}{m}$.Given, $(\frac{2}{t}) 	imes (\frac{2}{m}) = -1 \Rightarrow tm = -4$.The equation of the line $PQ$ passing through $(t^{2}, 2t)$ and $(m^{2}, 2m)$ is:$y - 2t = \frac{2m-2t}{m^{2}-t^{2}}(x - t^{2})$$y - 2t = \frac{2(m-t)}{(m-t)(m+t)}(x - t^{2})$$y - 2t = \frac{2}{m+t}(x - t^{2})$.Since $PQ$ intersects the axis of the parabola ($x$-axis) at $R(\alpha, 0)$, we substitute $y=0$ and $x=\alpha$:$0 - 2t = \frac{2}{m+t}(\alpha - t^{2})$$-t(m+t) = \alpha - t^{2}$$-tm - t^{2} = \alpha - t^{2}$$\alpha = -tm$.Since $tm = -4$, we have $\alpha = -(-4) = 4$.Thus, the distance of the vertex $X(0, 0)$ from $R(4, 0)$ is $4$.

Let $P$ and $Q$ be points on the parabola $y^{2}=4x$ such that the line segment $PQ$ subtends a right angle at the vertex. If $PQ$ intersects the axis of the parabola at $R$, then the distance of the vertex from $R$ is

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