Question

guided practice 144. fill in the table with the information needed to graph the equation $9y^2 - 16x^2 = 144$ equation of hyperbola $9y^2 - 16x^2 = 144$ standard form of the equation transverse axis values of a and b vertices dimensions of central rectangle what is the standard form of the equation $9y^2 - 16x^2 = 144$? a. $\frac{x^2}{9} - \frac{y^2}{16} = 1$ b. $\frac{y^2}{16} - \frac{x^2}{9} = 1$ c. $\frac{x^2}{16} - \frac{y^2}{9} = 1$ d. $\frac{y^2}{9} - \frac{x^2}{16} = 1$

Explanation:

Step1: Divide all terms by 144

$$\frac{9y^2}{144} - \frac{16x^2}{144} = \frac{144}{144}$$

Step2: Simplify each fraction

$$\frac{y^2}{16} - \frac{x^2}{9} = 1$$

Step1: Identify $a^2$ and $b^2$

From $\frac{y^2}{16} - \frac{x^2}{9} = 1$, $a^2=16$, $b^2=9$

Step2: Calculate $a$ and $b$

$a = \sqrt{16}=4$, $b = \sqrt{9}=3$

Step3: Determine transverse axis

Since $y^2$ term is positive, axis is vertical (y-axis)

Step4: Find vertices

Vertices: $(0, \pm a) = (0, 4), (0, -4)$

Step5: Central rectangle dimensions

Length: $2a=8$, Width: $2b=6$

Full Table Answers:

Category	Information
Transverse axis	Vertical (y-axis)
Values of $a$ and $b$	$a=4$, $b=3$
Vertices	$(0, 4)$, $(0, -4)$
Dimensions of central rectangle	$8 \times 6$

Answer:

B. $\frac{y^2}{16} - \frac{x^2}{9} = 1$

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Completion of the full table: