Answered: Graph the hyperbola…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Graph the hyperbola 16x^2−32x−4y^2−24y−84=0, noting the center, vertices, cover-tices, and foci.

Expert Solution

Step 1

The given equation of hyperbola is $16 x^{2} - 32 x - 4 y^{2} - 24 y - 84 = 0$ .

Convert the equation of hyperbola into its standard form as follows.

$16 x^{2} - 32 x - 4 y^{2} - 24 y - 84 = 0 16 x^{2} - 32 x - 4 y^{2} - 24 y = 84 4 x^{2} - 8 x - y^{2} - 6 y = 21 4 (x^{2} - 2 x) - (y^{2} + 6 y) = 21 4 (x^{2} - 2 x + 1) - (y^{2} + 6 y + 9) = 21 + 4 - 9 4 {(x - 1)}^{2} - {(y + 3)}^{2} = 16 \frac{{(x - 1)}^{2}}{4} - \frac{{(y + 3)}^{2}}{16} = 1$

Thus, the standard form of the given hyperbola is $\frac{{(x - 1)}^{2}}{4} - \frac{{(y + 3)}^{2}}{16} = 1$ .

Step 2

Sketch the graph of the hyperbola using any graphing calculator as shown below.

From the above figure, notice that (1, -3) is the center of the hyperbola.

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