Find the center verice foci and asyptote of the hyperbola?

Question

x^2/36 - y^2/64

 

Do i have to squre 36 and 64 to get the vertex and foci?

Answer 1

I will assume that the equation is:

 

x²/ 36 - y²/64  = 1         which is the same as   (x/6)² - (y/8)² =  1

 

The standard form for an East-West hyperbola is  (x/a)² - (y/b)² =1 

 

So this is an East-West hyperbola centered at the origin,  with a = 6 and b = 8. 

The vertex on  the right side is    (a,0) = (6,0)   

The distance from the center to the focal point is  c = sqrt(a² + b²)    = 10, so

the focal point on the right side is  (10,0)

 

The asymptotes cross at the origin.  The slopes are   ± b/a =  ± 4/3 .

 

The formula for c, the distance from center to focal point,  above should be memorized.  The other items can be easily figured out from the standard form equation.