The graph of the equation x^2/169-y^2/16=1 is a hyperbola.................5 stars best answer

the standard form equation of a hyperbola is x^2/a^2-y^2/b^2=1.
PART I : what are the values of a and Bin the given equation?
A=------------------------------------…
B=------------------------------------…

PART II: the slopes of the asymptotes for a hyperbola are b/a and -b/a. what are the slopes of the hyperbola's asymptotes using the given equation? show your work.

Part I:
Since 13^2=169 and 4^2=16, we see that a=13 and b=4. This makes our equation match the standard form perfectly.

Part II:
Since a=13 and b=4, we plug in these values to the formulas for the asymptotes' slopes and get them to be 4/13 and -4/13.

I.a is 13

b is 4

to find these the the square roots of the denominators

part II
4/13
and -4/13

these are simple the a an b from part 1. think of it as slope or "y/x"