Question

asked Dec 25, 2021 74.2k views

2 Answers

Kyre · Answer 1 · 2021-12-29T22:53:03+0000

Answer:

the positive slope of the asymptote = 5

Explanation:

Given that:

((y+11)^2)/(100) -((x-6)^2)/(4) =1

Using the standard form of the equation:

((y-k)^2)/(a^2)-((x-h)^2)/(b^2)= 1

where:

(h,k) are the center of the hyperbola.

and the y term is in front of the x term indicating that the hyperbola opens up and down.

a = distance that indicates how far above and below of the center the vertices of the hyperbola are.

For the above standard equation; the equation for the asymptote is:

$y = \pm (a)/(b) (x-h)+k$

where;

(a)/(b) is the slope

From above;

(h,k) = 11, 100

a^2 = 100

a =
√(100)

a = 10

b^2 =4

b =
√(4)

b = 2

$y = \pm (10)/(2) (x-11)+2$

$y = \pm 5 (x-11)+2$

y = 5x-53 , -5x -57

Since we are to find the positive slope of the asymptote: we have

(a)/(b) to be the slope in the equation
$y = \pm (10)/(2) (y-11)+2$

(a)/(b) =
(10)/(2)

(a)/(b) = 5

Thus, the positive slope of the asymptote = 5

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