Quiz about Hyperbola Quiz 10.3

Question 1

Write in standard hyperbolic form, $$x^2-4y^2+10x+32y-75=0$$ :

Accepted Answer

$$\frac{(x+5)^2}{36}-\frac{(y-4)^2}{9}=1$$

Answer

$$\frac{\left(x+5\right)^2}{164}-\frac{\left(y-4\right)^2}{41}=1$$

Answer

$$\frac{\left(x+5\right)^2}{36}-\frac{\left(y+4\right)^2}{9}=1$$

Answer

$$\frac{(y-4)^2}{9}-\frac{(x-5)^2}{36}=1$$

Question 2

What are the equations of the asymptotes of the hyperbola in the picture?

Accepted Answer

y = 2x and y = -2x

Answer

y = x and y = -x

Answer

y = ¹/₂x and y = -¹/₂x

Answer

y = 2x and y = 4x

Question 3

Find the equation of the hyperbola.

Accepted Answer

$$\frac{(x-2)^2}{16}-\frac{(y+3)^2}{20}=1$$

Answer

$$\frac{(x+3)^2}{16}-\frac{(y-2)}{20}=1$$

Answer

$$\frac{(x-2)^2}{20}-\frac{(y+3)^2}{16}=1$$

Answer

$$\frac{(y+3)^2}{16}-\frac{(x-2)^2}{20}=1$$

Question 4

What is the center of the hyperbola?

$$\frac{y^2}{9}-(x+4)^2=1$$

Accepted Answer

(-4, 0)

Answer

(0, 4)

Answer

(4, 0)

Answer

(0, -4)

Question 5

What is the correct equation?

Accepted Answer

$$\frac{y^2}{25}-\frac{x^2}{9}=1$$

Answer

$$\frac{\left(x-1\right)^2}{16}-\frac{y^2}{4}=1$$

Answer

$$\frac{\left(x+1\right)^2}{16}-\frac{\left(y-2\right)^2}{9}=1$$

Answer

$$\frac{y^2}{9}-\frac{x^2}{25}=1$$

Question 6

Write the standard form of the following equation:

x² - 4y² + 6x - 8y = 11

Accepted Answer

$$\frac{\left(x+3\right)^2}{16}-\frac{\left(y+1\right)^2}{4}=1$$

Answer

$$\frac{\left(x+3\right)^2}{16}-\frac{\left(y-1\right)^2}{4}=1$$

Answer

$$\left(x+3\right)^2-\frac{\left(y+1\right)^2}{4}=1$$

Answer

$$\frac{\left(x+3\right)^2}{16}-\left(y+1\right)^2=1$$

Hyperbola Quiz 10.3