Question:

For the parabola $y^2 = 8x$, we have $a = 2$. The general equation of a tangent with slope $m$ (or parameter $t$) is: $$ty = x + 2t^2$$ Since both tangents pass through $C_1(-4, 0)$: $$0 = -4 + 2t^2 \implies t^2 = 2 \implies t = \pm\sqrt{2}$$ Thus, the points of contact $A_1$ and $B_1$ are: $$A_1 = (at^2, 2at) = (4, 4\sqrt{2}) \quad \text{and} \quad B_1 = (4, -4\sqrt{2})$$ Let's verify the statements: - Option A: $OA_1 = \sqrt{4^2 + (4\sqrt{2})^2} = \sqrt{16+32} = 4\sqrt{3}$ (Option A is True). - Option B: $A_1B_1 = 4\sqrt{2} - (-4\sqrt{2}) = 8\sqrt{2} \neq 16$ (Option B is False). - Option C/D: The vertices of $\triangle A_1B_1C_1$ are $A_1(4, 4\sqrt{2})$, $B_1(4, -4\sqrt{2})$, and $C_1(-4, 0)$. - Since $A_1B_1$ is a vertical line segment (on $x = 4$), the altitude from $C_1$ is horizontal, which lies on the line $y = 0$ (the $x$-axis). - The slope of side $B_1C_1$ is: $$m_{B_1C_1} = \dfrac{-4\sqrt{2} - 0}{4 - (-4)} = -\dfrac{1}{\sqrt{2}}$$ - The altitude from $A_1$ to $B_1C_1$ has slope $m_{alt} = \sqrt{2}$. Its equation is: $$y - 4\sqrt{2} = \sqrt{2}(x - 4) \implies y = \sqrt{2}x$$ - The intersection of the two altitudes ($y = 0$ and $y = \sqrt{2}x$) is $(0,0)$. Thus, the orthocenter is $(0,0)$ (Option C is True, D is False). Hence, the correct options are A and C.