Let y=mx+c, m0 be the focal chord of y^{2}=- | vedclass

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(A) The equation of the parabola is $y^{2}=-64x$. Comparing with $y^{2}=-4ax$,we get $a=16$. The focus is $(-16, 0)$.Since $y=mx+c$ is a focal chord,i

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$34$

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(A) The equation of the parabola is $y^{2}=-64x$. Comparing with $y^{2}=-4ax$,we get $a=16$. The focus is $(-16, 0)$.Since $y=mx+c$ is a focal chord,it passes through $(-16, 0)$,so $0 = m(-16) + c$,which gives $c=16m$.The line $y=mx+c$ is tangent to the circle $(x+10)^{2}+y^{2}=4$. The center of the circle is $(-10, 0)$ and the radius is $r=2$.The perpendicular distance from the center $(-10, 0)$ to the line $mx-y+c=0$ is equal to the radius $r=2$.$\frac{|m(-10)-0+c|}{\sqrt{m^{2}+(-1)^{2}}} = 2$$|c-10m| = 2\sqrt{m^{2}+1}$.Substituting $c=16m$,we get $|16m-10m| = 2\sqrt{m^{2}+1}$,so $|6m| = 2\sqrt{m^{2}+1}$.Since $m>0$,$3m = \sqrt{m^{2}+1}$. Squaring both sides,$9m^{2} = m^{2}+1$,so $8m^{2}=1$,which gives $m=\frac{1}{2\sqrt{2}}$.Then $c = 16m = 16 	imes \frac{1}{2\sqrt{2}} = \frac{8}{\sqrt{2}}$.Finally,$4\sqrt{2}(m+c) = 4\sqrt{2}(\frac{1}{2\sqrt{2}} + \frac{8}{\sqrt{2}}) = 4\sqrt{2}(\frac{1+16}{2\sqrt{2}}) = 2(17) = 34$.

Let $y=mx+c, m>0$ be the focal chord of $y^{2}=-64x$,which is tangent to $(x+10)^{2}+y^{2}=4$. Then,the value of $4\sqrt{2}(m+c)$ is equal to $.....$

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