SOLUTION: Find the verticies of the hyperboloa defined by this equation: (x-2)^2 over 36 minus (y-2)^2 over 4 equals 1.

Algebra.Com
Question 614581: Find the verticies of the hyperboloa defined by this equation:
(x-2)^2 over 36 minus (y-2)^2 over 4 equals 1.
Found 2 solutions by lwsshak3, ewatrrr:
Answer by lwsshak3(11628)   (Show Source): You can put this solution on YOUR website!
Find the verticies of the hyperboloa defined by this equation:
(x-2)^2 over 36 minus (y-2)^2 over 4 equals 1
**
(x-2)^2/36-(y-2)^2/4=1
This is an equation of a hyperbola with horizontal transverse axis.
Its standard form: (x-h)^2/a^2-(y-k)^2/b^2=1, (h,k)=(x,y) coordinates of center
For given equation: (x-2)^2/36-(y-2)^2/4=1
center: (2,2)
a^2=36
a=√36=6
vertices: (2±a,2)=(2±6,2)=(-4,2) and (8,2)
Answer by ewatrrr(24785)   (Show Source): You can put this solution on YOUR website!
 

Hi,

 Opens right and left along y = 2   C(2,2) 

  

  Foci( (2-sqrt(40), 2) and (2+sqrt(40), 2) 

RELATED QUESTIONS


the process is elimination by substitution. Equation(1) x+y over 2 minus x-y over 3... (answered by edjones)

Find the vertices and asymptotes of the hyperbola given by
x^2/25-y^2/2^2=1
(x squared... (answered by jsmallt9)

Solve

5 over 3 minus 2 over x-4 equals 1 over... (answered by stanbon)

Graph the conic:
(x-2)^2 over 36 + (y+4)^2 over 25 =... (answered by ewatrrr)

Solve using quadratic formula.

1 over X^2 minus 3 equals 8 over x

The solution is... (answered by Theo,MathTherapy)

Find the value of x in each proportion. 
a) x + 1 over 3 equals 10 over x + 2
b) x - 2... (answered by stanbon)

Not sure if this is the appropriate topic

Directions says to solve

x minus 1 over x  (answered by solver91311)

  what are the asymptotes of (y+3^2) over (36) − (x+4^2) over (64)... (answered by Fombitz)

please help me with this problem.  I haven't been able to find the answer.
thank you... (answered by tanimachatterjee)