Question

What is the positive slope of the asymptote of (y+11)^2/100-(x-6)^2/4=1? The positive slope of the asymptote is .

Answer 1

Answer:

the  positive slope of the asymptote = 5

Step-by-step explanation:

Given that:

$\frac{(y+11)^2}{100} -\frac{(x-6)^2}{4} =1$

Using the standard form of the equation:

$\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}= 1$

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:

$y = \pm \frac{a}{b} (x-h)+k$

where;

$\frac{a}{b}$ is the slope

From above;

(h,k) = 11, 100

$a^2$ = 100

a = $\sqrt{100}$

a = 10

$b^2 =4$

b = $\sqrt{4}$

b = 2

$y = \pm \frac{10}{2} (x-11)+2$

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

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

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

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

$\frac{a}{b}$  = $\frac{10}{2}$

$\frac{a}{b}$  = 5

Thus, the  positive slope of the asymptote = 5