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(B) For the first curve,$x = t^2 + 1$ and $y = 2t$. Substituting $t = \frac{y}{2}$ into the equation for $x$,we get $x = (\frac{y}{2})^2 + 1$,which si

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$(2, 2)$

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(B) For the first curve,$x = t^2 + 1$ and $y = 2t$. Substituting $t = \frac{y}{2}$ into the equation for $x$,we get $x = (\frac{y}{2})^2 + 1$,which simplifies to $x = \frac{y^2}{4} + 1$,or $y^2 = 4(x - 1)$.For the second curve,$x = 2s$ and $y = \frac{2}{s}$. Multiplying these gives $xy = (2s)(\frac{2}{s}) = 4$,so $xy = 4$.To find the intersection,substitute $y = \frac{4}{x}$ into $y^2 = 4x - 4$:$(\frac{4}{x})^2 = 4x - 4$$\frac{16}{x^2} = 4(x - 1)$$4 = x^2(x - 1)$$x^3 - x^2 - 4 = 0$.Testing $x = 2$,we get $8 - 4 - 4 = 0$,which is true.If $x = 2$,then $y = \frac{4}{2} = 2$.Thus,the point of intersection is $(2, 2)$.

The points of intersection of the curves whose parametric equations are $x = t^2 + 1, y = 2t$ and $x = 2s, y = \frac{2}{s}$ is given by

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