Question

What are the vertices, foci, and asymptotes of the hyperbola with the equation 16x^2 – 4y^2 = 64?

Answer 1

when x=0, there is not value of y that can create a positive 64; this tells us that the verts and foci are of the form (x,0)

Answer 2

\[\frac{(nx)^2}{1}-\frac{(my)^2}{1}=r^2\]
\[\frac{(nx)^2}{(r)^2}-\frac{(my)^2}{(r)^2}=1\]
\[\frac{x^2}{(r/n)^2}-\frac{y^2}{(r/m)^2}=1\]
this might help to see how things play out

Answer 3

\[16x^2 – 4y^2 = 64\]
\[\frac{16x^2}{64} – \frac{4y^2}{64} = 1\]
\[\frac{x^2}{64/16} – \frac{y^2}{64/4} = 1\]
\[\frac{x^2}{4} – \frac{y^2}{16} = 1\]
\[\frac{x^2}{2^2} – \frac{y^2}{16^2} = 1\]

Answer 4

so tell me, when y=0, what values of x make this a true statement? those will define you vertexes

Answer 5

Alright, I found the answers a few seconds ago and scored a 100% thanks so much for the help you gave!

Answer 6

youre welcome, good luck