If Q is the point on the parabola y^2=4x that | vedclass

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(B) Let the coordinates of point $Q$ be $(x, y)$ on the parabola $y^2=4x$. The distance $PQ$ is given by $PQ = \sqrt{(x-2)^2 + (y-0)^2}$. Substituting

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$2$

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(B) Let the coordinates of point $Q$ be $(x, y)$ on the parabola $y^2=4x$. The distance $PQ$ is given by $PQ = \sqrt{(x-2)^2 + (y-0)^2}$. Substituting $y^2 = 4x$, we get $PQ = \sqrt{(x-2)^2 + 4x} = \sqrt{x^2 - 4x + 4 + 4x} = \sqrt{x^2 + 4}$. To find the minimum distance, let $f(x) = x^2 + 4$. Differentiating with respect to $x$, $f'(x) = 2x$. Setting $f'(x) = 0$, we get $x = 0$. Since $f''(x) = 2 > 0$, the function has a minimum at $x = 0$. The minimum distance is $PQ = \sqrt{0^2 + 4} = \sqrt{4} = 2$.

If $Q$ is the point on the parabola $y^2=4x$ that is nearest to the point $P(2,0)$, then $PQ=$

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