The focal chord to y^2 = 16x is tangent to (x | vedclass

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(A) The parabola is $y^2 = 16x$,so $4a = 16$,which gives $a = 4$. The focus is $(a, 0) = (4, 0)$.Any focal chord passing through $(4, 0)$ with slope $

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$\{-1, 1\}$

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(A) The parabola is $y^2 = 16x$,so $4a = 16$,which gives $a = 4$. The focus is $(a, 0) = (4, 0)$.Any focal chord passing through $(4, 0)$ with slope $m$ has the equation $y - 0 = m(x - 4)$,or $mx - y - 4m = 0$.This line is tangent to the circle $(x - 6)^2 + y^2 = 2$,which has center $(6, 0)$ and radius $r = \sqrt{2}$.The perpendicular distance from the center $(6, 0)$ to the line $mx - y - 4m = 0$ must be equal to the radius $\sqrt{2}$.Using the distance formula: $\frac{|m(6) - 0 - 4m|}{\sqrt{m^2 + (-1)^2}} = \sqrt{2}$.$\frac{|2m|}{\sqrt{m^2 + 1}} = \sqrt{2}$.Squaring both sides: $\frac{4m^2}{m^2 + 1} = 2$.$4m^2 = 2m^2 + 2$.$2m^2 = 2$,so $m^2 = 1$,which gives $m = \pm 1$.Thus,the possible values of the slope are $\{-1, 1\}$.

The focal chord to $y^2 = 16x$ is tangent to $(x - 6)^2 + y^2 = 2$. Then,the possible values of the slope of this chord are:

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