Let P be a point on the hyperbola x^2 - y^2 = | vedclass

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(B) Let $P(x, y)$ be a point on the hyperbola $x^2 - y^2 = 4$,so $x^2 = y^2 + 4$.Let $Q$ be the point $(0, -1)$. The distance $PQ$ is given by $PQ^2 =

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$\frac{1}{2}$

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(B) Let $P(x, y)$ be a point on the hyperbola $x^2 - y^2 = 4$,so $x^2 = y^2 + 4$.Let $Q$ be the point $(0, -1)$. The distance $PQ$ is given by $PQ^2 = (x - 0)^2 + (y - (-1))^2 = x^2 + (y + 1)^2$.Substituting $x^2 = y^2 + 4$,we get $PQ^2 = y^2 + 4 + y^2 + 2y + 1 = 2y^2 + 2y + 5$.To minimize $PQ^2$,we differentiate with respect to $y$: $f(y) = 2y^2 + 2y + 5$.$f'(y) = 4y + 2 = 0 \implies y = -\frac{1}{2}$.The distance of $P$ from the $x$-axis is $|y| = |-\frac{1}{2}| = \frac{1}{2}$.

Let $P$ be a point on the hyperbola $x^2 - y^2 = 4$,which is at the minimum distance from $(0, -1)$. Then the distance of $P$ from the $x$-axis is:

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