Chapter 9: Q. 39 (page 772)

Use Cartesian coordinates to express the equations for the hyperbolas determined by the conditions specified in Exercises 38–43.
$foci (\pm 6, 0), directrices x = \pm 1$

Short Answer

Expert verified

The equation is $\frac{x^{2}}{6} - \frac{y^{2}}{30} = 1 .$

Step by step solution

Step 1. Given information.

The given values are,

$foci (\pm 6, 0), directrices x = \pm 1$

Step 2. Value of variables.

Now,

$The focus points are (6, 0), (- 6, 0) . Center = (\frac{6 - 6}{2}, \frac{0 + 0}{2}) [since mid point = (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})] Center = (0, 0) Given driectries are x = \pm 1 . a = e Then a e = 6 a^{2} = 6 c = \sqrt{(0 - 6)^{2} + (0 - 0)^{2}} [since D = \sqrt{{(x_{2} - x_{1})}^{2} + {(y_{2} - y_{1})}^{2}}] c = 6 For a hyperbola, a^{2} + b^{2} = c^{2} 6 + b^{2} = 6^{2} b^{2} = 30$

Step 3. Substitution.

Now, substitute the obtained values in,

$\frac{(x - h)^{2}}{a^{2}} - \frac{(y - k)^{2}}{b^{2}} = 1 where (h, k) is the center. \frac{(x - 0)^{2}}{6} - \frac{(y - 0)^{2}}{30} = 1 [since a^{2} = 6, b^{2} = 30] \frac{x^{2}}{6} - \frac{y^{2}}{30} = 1$

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Use Cartesian coordinates to express the equations for the hyperbolas determined by the conditions specified in Exercises 38–43.
$foci (\pm 6, 0), directrices x = \pm 1$

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Value of variables.

Step 3. Substitution.

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Use Cartesian coordinates to express the equations for the hyperbolas determined by the conditions specified in Exercises 38–43.foci(±6,0), directricesx=±1

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Value of variables.

Step 3. Substitution.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Calculus

Mechanics Maths

Decision Maths

Probability and Statistics

Applied Mathematics

Pure Maths

Study anywhere. Anytime. Across all devices.

Use Cartesian coordinates to express the equations for the hyperbolas determined by the conditions specified in Exercises 38–43.
$foci (\pm 6, 0), directrices x = \pm 1$