Let A_1, B_1, C_1 be three points in the xy-p | vedclass

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(A,C) The equation of the tangent to the parabola $y^2=8x$ (where $4a=8$,so $a=2$) at point $(2t^2, 4t)$ is given by $ty = x + 2t^2$.Since the tangent

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$A, C$

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(A,C) The equation of the tangent to the parabola $y^2=8x$ (where $4a=8$,so $a=2$) at point $(2t^2, 4t)$ is given by $ty = x + 2t^2$.Since the tangents pass through $C_1 = (-4, 0)$,we have $0 = -4 + 2t^2$,which implies $t^2 = 2$,so $t = \pm \sqrt{2}$.Thus,the points of tangency are $A_1 = (2(\sqrt{2})^2, 4\sqrt{2}) = (4, 4\sqrt{2})$ and $B_1 = (2(-\sqrt{2})^2, 4(-\sqrt{2})) = (4, -4\sqrt{2})$.$(A)$ The length of $OA_1 = \sqrt{4^2 + (4\sqrt{2})^2} = \sqrt{16 + 32} = \sqrt{48} = 4\sqrt{3}$. Thus,$(A)$ is $TRUE$.$(B)$ The length of $A_1 B_1 = |4\sqrt{2} - (-4\sqrt{2})| = 8\sqrt{2}$. Thus,$(B)$ is $FALSE$.$(C)$ The slope of $A_1 B_1$ is undefined (vertical line $x=4$). The altitude from $C_1(-4, 0)$ to $A_1 B_1$ is the horizontal line $y=0$.The slope of $A_1 C_1$ is $\frac{4\sqrt{2}-0}{4-(-4)} = \frac{4\sqrt{2}}{8} = \frac{1}{\sqrt{2}}$. The altitude from $B_1(4, -4\sqrt{2})$ to $A_1 C_1$ has slope $-\sqrt{2}$.The equation of this altitude is $y - (-4\sqrt{2}) = -\sqrt{2}(x - 4) \Rightarrow y + 4\sqrt{2} = -\sqrt{2}x + 4\sqrt{2} \Rightarrow y = -\sqrt{2}x$.Intersection of $y=0$ and $y=-\sqrt{2}x$ is $(0,0)$. Thus,$(C)$ is $TRUE$.

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