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

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(A) The equation of the parabola is $y^2 = 16x$,so $4a = 16$,which gives $a = 4$. The focus of the parabola is $(a, 0) = (4, 0)$. The equation of a fo

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$\{ -1, 1 \}$

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(A) The equation of the parabola is $y^2 = 16x$,so $4a = 16$,which gives $a = 4$. The focus of the parabola is $(a, 0) = (4, 0)$. The equation of a focal chord with slope $m$ passing through $(4, 0)$ is $y - 0 = m(x - 4)$,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 $\sqrt{2}$. $\frac{|m(6) - 0 - 4m|}{\sqrt{m^2 + (-1)^2}} = \sqrt{2}$ $\frac{|2m|}{\sqrt{m^2 + 1}} = \sqrt{2}$ Squaring both sides,we get $\frac{4m^2}{m^2 + 1} = 2$. $4m^2 = 2m^2 + 2$ $2m^2 = 2$ $m^2 = 1$ $m = \pm 1$.

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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