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Exercises Conic sections

Example 1 from Algebra and Trigonometry, 12.3 The Parabola

Graphing a Parabola with Vertex (0, 0) and the x-axis as the Axis of Symmetry

Graph $y^{2} = 24 x .$ Identify and label the focus, directrix, and endpoints of the latus rectum.

Work it out on paper first. Then open the solution one step at a time, and stop as soon as you can finish on your own.

1
Solution
The standard form that applies to the given equation is $y^{2} = 4 p x .$ Thus, the axis of symmetry is the x-axis. It follows that:
- $24 = 4 p,$ so $p = 6.$ Since $p > 0,$ the parabola opens right
- the coordinates of the focus are $(p, 0) = (6, 0)$
- the equation of the directrix is $x = - p = - 6$
- the endpoints of the latus rectum have the same x-coordinate at the focus. To find the endpoints, substitute $x = 6$ into the original equation: $(6, \pm 12)$
Next we plot the focus, directrix, and latus rectum, and draw a smooth curve to form the parabola. Figure 7

Figure 7

Your turn

The book's Try It right after this example: same idea, new numbers. Only the answer is given.

Graph $y^{2} = −16 x .$ Identify and label the focus, directrix, and endpoints of the latus rectum.

How did it go?

PreviousSolving Applied Problems Involving Hyperbolas Next Graphing a Parabola with Vertex (0, 0) and the y-axis as the Axis of Symmetry

From Algebra and Trigonometry (1st edition) by OpenStax (Rice University), section 12.3, licensed under CC BY 4.0. Original at openstax.org. We split the solution into steps; the words, numbers and figures are the book's. OpenStax doesn't endorse this site.