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Ch. 7 - Conic Sections
Blitzer - College Algebra 8th Edition
Blitzer8th EditionCollege AlgebraISBN: 9780136970514Not the one you use?Change textbook
Chapter 8, Problem 5

Find the focus and directrix of the parabola with the given equation. Then graph the parabola. y2=16xy^2 = 16x

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1
Recognize that the given equation \(y^2 = 16x\) is a parabola that opens horizontally because the \(y\) variable is squared and \(x\) is to the first power.
Rewrite the equation in the standard form of a horizontal parabola: \(y^2 = 4px\). Here, \(4p = 16\), so solve for \(p\) by dividing both sides by 4, giving \(p = 4\).
Identify the vertex of the parabola, which is at the origin \((0,0)\) since the equation is not shifted.
Determine the focus using the value of \(p\). For a parabola that opens to the right, the focus is at \((p, 0)\), so substitute \(p = 4\) to get the focus at \((4, 0)\).
Find the directrix, which is a vertical line given by \(x = -p\). Substitute \(p = 4\) to get the directrix equation \(x = -4\).

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Standard Form of a Parabola

The standard form of a parabola's equation helps identify its orientation and key features. For a parabola opening horizontally, the form is (y - k)^2 = 4p(x - h), where (h, k) is the vertex and p determines the distance to the focus and directrix.
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Parabolas as Conic Sections

Focus and Directrix of a Parabola

The focus is a fixed point inside the parabola, and the directrix is a line outside it. The parabola consists of points equidistant from the focus and directrix. The value p in the equation gives the distance from the vertex to both the focus and directrix.
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Horizontal Parabolas

Graphing a Parabola

Graphing involves plotting the vertex, focus, and directrix, then sketching the curve that is symmetric about the axis through the vertex and focus. Knowing the orientation (horizontal or vertical) and distances helps accurately draw the parabola.
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Horizontal Parabolas