Question

A hyperbola is represented using the equation (x-1)^2/4-(y+2)^2/16=1. What are the slopes of the asymptotes?

Answer 1

Answer:

the slope of the asymptotes are: 2 and -2

Step-by-step explanation:

As we know the hyperbola equation is:

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

Given the hyperbola equation:

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

We have  a=2 and b=4, h=1 and k=-2

and the asymptote equation of a translate hyperbola with equation is:

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

We substitute this values to get:

$y=\pm\frac{4}{2}(x-1) -2$  

<=> y = 2x -4 or y = -2x

so the slope of the asymptotes are: 2 and -2