Question

Find the vertex, focus, and directrix of the parabola with the given equation. Then graph the parabola. $$(x-4)^2 = 4(y+1)$$

Accepted Answer

Rewrite the given equation in standard form for a parabola. The equation $$(x-h)^2 = 4p(y-k)$$ represents a parabola that opens vertically, where $$(h, k) is $$the vertex and $$p$$ determines the distance from the vertex to the focus and directrix. In this case, $$(x-4)^2 = 4(y+1) is $$already in standard form with $$h = 4$$, $$k = -1$$, and $$4p = 4. $$Identify the vertex of the parabola. From the equation, the vertex is $$(h, k) = (4, -1). $$Determine the value of $$p. $$Since $$4p = 4$$, divide both sides by 4 to find $$p = 1. $$This means the focus is 1 unit away from the vertex along the axis of symmetry, and the directrix is 1 unit away in the opposite direction. Find the focus. Since the parabola opens vertically, the focus is located $$p$$ units above the vertex. Add $$p = 1 to $$the $$y-$$coordinate of the vertex: $$(4, -1 + 1) = (4, 0). $$Find the directrix. The directrix is a horizontal line $$p$$ units below the vertex. Subtract $$p = 1$$ from the $$y-$$coordinate of the vertex: $$y = -1 - 1 = -2. $$The directrix is the line $$y = -2.$$

Find the vertex, focus, and directrix of the parabola with the given equation. Then graph the parabola. (x-4)^2 = 4(y+1)

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Key Concepts

Parabola Definition

Vertex of a Parabola

Focus and Directrix