Answer:
the positive slope of the asymptote = 5
Step-by-step explanation:
Given that:
[tex]\frac{(y+11)^2}{100} -\frac{(x-6)^2}{4} =1[/tex]
Using the standard form of the equation:
[tex]\frac{(y-k)^2}{a^2}-\frac{(x-h)^2}{b^2}= 1[/tex]
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:
[tex]y = \pm \frac{a}{b} (x-h)+k[/tex]
where;
[tex]\frac{a}{b}[/tex] is the slope
From above;
(h,k) = 11, 100
[tex]a^2[/tex] = 100
a = [tex]\sqrt{100}[/tex]
a = 10
[tex]b^2 =4[/tex]
b = [tex]\sqrt{4}[/tex]
b = 2
[tex]y = \pm \frac{10}{2} (x-11)+2[/tex]
[tex]y = \pm 5 (x-11)+2[/tex]
y = 5x-53 , -5x -57
Since we are to find the positive slope of the asymptote: we have
[tex]\frac{a}{b}[/tex] to be the slope in the equation [tex]y = \pm \frac{10}{2} (y-11)+2[/tex]
[tex]\frac{a}{b}[/tex] = [tex]\frac{10}{2}[/tex]
[tex]\frac{a}{b}[/tex] = 5
Thus, the positive slope of the asymptote = 5
