Example 5 from Algebra and Trigonometry, 12.3 The Parabola

Graphing a Parabola from an Equation Given in General Form

Graph $x^{2} - 8 x - 28 y - 208 = 0.$ Identify and label the vertex, axis of symmetry, 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
Start by writing the equation of the parabola in standard form. The standard form that applies to the given equation is ${(x - h)}^{2} = 4 p (y - k) .$ Thus, the axis of symmetry is parallel to the y-axis. To express the equation of the parabola in this form, we begin by isolating the terms that contain the variable $x$ in order to complete the square.

$\begin{array}{l} x^{2} - 8 x - 28 y - 208 = 0 \\ x^{2} - 8 x = 28 y + 208 \\ x^{2} - 8 x + 16 = 28 y + 208 + 16 \\ {(x - 4)}^{2} = 28 y + 224 \\ {(x - 4)}^{2} = 28 (y + 8) \\ {(x - 4)}^{2} = 4 \cdot 7 \cdot (y + 8) \end{array}$
It follows that:
- the vertex is $(h, k) = (4, −8)$
- the axis of symmetry is $x = h = 4$
- since $p = 7, p > 0$ and so the parabola opens up
- the coordinates of the focus are $(h, k + p) = (4, −8 + 7) = (4, −1)$
- the equation of the directrix is $y = k - p = −8 - 7 = −15$
- the endpoints of the latus rectum are $(h \pm 2 p, k + p) = (4 \pm 2 (7), −8 + 7),$ or $(−10, −1)$ and $(18, −1)$
Next we plot the vertex, axis of symmetry, focus, directrix, and latus rectum, and draw a smooth curve to form the parabola. See Figure 11.

Figure 11

Your turn

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

Graph ${(x + 2)}^{2} = −20 (y - 3) .$ Identify and label the vertex, axis of symmetry, focus, directrix, and endpoints of the latus rectum.

How did it go?