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For the parabola $y^2 = 16x$, find the equation of the directrix.
A$x = -4$
B$x = 4$
C$y = -4$
D$y = 4$
Answer & Solution
Correct answer: A. $x = -4$
**Identify $a$.** $y^2 = 4ax$ with $4a = 16 \Rightarrow a = 4$.
**Locate the directrix.** For $y^2 = 4ax$ the directrix is the vertical line $x = -a$ (on the opposite side of the vertex from the focus). So $x = -4$.
**Geometric sanity check.** Pick any point on $y^2 = 16x$, say $(1, 4)$. Distance to focus $(4, 0)$: $\sqrt{9 + 16} = 5$. Distance to directrix $x = -4$: horizontal $|1 - (-4)| = 5$. ✓ Equal — confirming the directrix.
**Trap options.** $x = 4$ (B) is the focus's $x$-coordinate, not the directrix. Options C and D ($y = \pm 4$) put a *horizontal* directrix, which is wrong for an axis-on-$x$ parabola.
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