The distance of the point (6, -2 sqrt{2}) fro | vedclass

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(B) For the parabola $y^2 = \frac{x}{2}$,the tangent line is $y = mx + \frac{1}{8m}$.For the curve $x = 1 + y^2$,substituting the tangent line gives $

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$5$

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(B) For the parabola $y^2 = \frac{x}{2}$,the tangent line is $y = mx + \frac{1}{8m}$.For the curve $x = 1 + y^2$,substituting the tangent line gives $x = 1 + (mx + \frac{1}{8m})^2$.Expanding this,$x = 1 + m^2x^2 + \frac{x}{4} + \frac{1}{64m^2}$,which simplifies to $m^2x^2 - \frac{3}{4}x + (1 + \frac{1}{64m^2}) = 0$.Since the line is a tangent,the discriminant $D = 0$,so $(-\frac{3}{4})^2 - 4(m^2)(1 + \frac{1}{64m^2}) = 0$.$\frac{9}{16} - 4m^2 - \frac{1}{16} = 0$ $\Rightarrow \frac{8}{16} = 4m^2$ $\Rightarrow m^2 = \frac{1}{8}$ $\Rightarrow m = \frac{1}{2\sqrt{2}}$ (since $m > 0$).Substituting $m$ into the tangent equation: $y = \frac{1}{2\sqrt{2}}x + \frac{1}{8(\frac{1}{2\sqrt{2}})} = \frac{1}{2\sqrt{2}}x + \frac{\sqrt{2}}{2} = \frac{1}{2\sqrt{2}}x + \frac{1}{\sqrt{2}}$.Multiplying by $2\sqrt{2}$,we get $2\sqrt{2}y = x + 2$,or $x - 2\sqrt{2}y + 2 = 0$.Wait,re-evaluating the constant: $c = \frac{1}{8m} = \frac{1}{8(\frac{1}{2\sqrt{2}})} = \frac{2\sqrt{2}}{8} = \frac{\sqrt{2}}{4} = \frac{1}{2\sqrt{2}}$.So the tangent is $x - 2\sqrt{2}y + 1 = 0$.The distance from $(6, -2\sqrt{2})$ to $x - 2\sqrt{2}y + 1 = 0$ is $d = \frac{|6 - 2\sqrt{2}(-2\sqrt{2}) + 1|}{\sqrt{1^2 + (-2\sqrt{2})^2}} = \frac{|6 + 8 + 1|}{\sqrt{1 + 8}} = \frac{15}{3} = 5$.

The distance of the point $(6, -2 \sqrt{2})$ from the common tangent $y = mx + c$ $(m > 0)$ of the curves $x = 2y^2$ and $x = 1 + y^2$ is

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