Math · Precalculus · Worked example
Find the focus and directrix of a parabola
Find the vertex, focus and directrix of y² = 12x, and check the focus–directrix property at the point (3, 6).
Match the standard form
y² = 4px with 4p = 12, so p = 3. The squared variable is y, so the parabola opens along the x-axis, to the right because p > 0.
Place the focus and directrix
The vertex is (0, 0). The focus is p units inside the curve, at (3, 0), and the directrix is the vertical line x = −3, p units the other way.
Check a point
(3, 6) is on the parabola, because 6² = 36 = 12 · 3. Its distance to the focus (3, 0) is 6, and its distance to the line x = −3 is 3 − (−3) = 6: equal, as the definition requires.
Result
Vertex (0, 0), focus (3, 0), directrix x = −3.
Your turn
Find the focus and directrix of x² = −8y.
Show the answer and explanation
Focus (0, −2); directrix y = 2.
x² = 4py with 4p = −8, so p = −2 and the parabola opens downward.
Keep exploring
In Graph, the upper half y = √(12x) passes through (3, 6). Change 12 to 4 in both halves and p drops to 1: the parabola narrows around its new focus, (1, 0).
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