SOLUTION: Can you help me with these? Please Find the standard form of the equation of the hyperbola with; 1. F_1 (5,0), F_2 (-5,0) V_1 (4,0), V_2 (-4,0)

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Can you help me with these? Please Find the standard form of the equation of the hyperbola with; 1. F_1 (5,0), F_2 (-5,0) V_1 (4,0), V_2 (-4,0)      Log On


   



Question 1185036: Can you help me with these? Please
Find the standard form of the equation of the hyperbola with;
1. F_1 (5,0), F_2 (-5,0)
V_1 (4,0), V_2 (-4,0)

Found 2 solutions by MathLover1, Edwin McCravy:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
the standard form of the equation of the hyperbola with;
1.
F%5B1%5D (5,0), F%5B2%5D (-5,0)
F%5B1%5D (c,0)=>c=5
V%5B1%5D (4,0), V%5B2%5D (-4,0)
V%5B1%5D (a,0) =>a=4
b%5E2=c%5E2-a%5E2
b%5E2=5%5E2-4%5E2
b%5E2=25-16
b%5E2=9
=>b=3
=> Co-vertices: (0,-3), (0,3)
=> center is at origin
Major (transverse) axis length: distance between vertices2a=8+
Semi-major axis length: a=4
Minor (conjugate) axis length: 2b=6
Semi-minor axis length:+b=3
equation of the hyperbola:
x%5E2%2F4%5E2-y%5E2%2F3%5E2=1
x%5E2%2F16-y%5E2%2F9=1





Answer by Edwin McCravy(20059) About Me  (Show Source):
You can put this solution on YOUR website!
We plot those given points and sketch in a rough sketch of the
hyperbola.  All "conic" graphs curve around their focal points.



The equation of all hyperbolas that open right and left have equations,

%28x-h%29%5E2%2Fa%5E2%22%22-%22%22%28y-k%29%5E2%2Fb%5E2%22%22=%22%221

where (h,k) is the center, "a" is the distance between the center and
either vertex. "c" is the distance between the center and either focal
point (or focus).  But "c" does not appear in the standard equation of
a hyperbola. We calculate "b2" by the Pythagorean theorem:

c%5E2=a%5E2%2Bb%5E2
5%5E2=4%5E2%2Bb%5E2
25=16%2Bb%5E2
9=b%5E2

The center is (0,0), so h=0, k=0, a2=4^2=16, b2=3^2=9.

Substitute in

%28x-h%29%5E2%2Fa%5E2%22%22-%22%22%28y-k%29%5E2%2Fb%5E2%22%22=%22%221

%28x-0%29%5E2%2F16%5E%22%22%22%22-%22%22%28y-0%29%5E2%2F9%5E%22%22%22%22=%22%221

x%5E2%2F16%5E%22%22%22%22-%22%22y%5E2%2F9%5E%22%22%22%22=%22%221

Edwin