|
Question 607294: Find the equation in standard form of a hyperbola with vertices (1,-3) and (-5,-3 and eccentricity 5/3
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Find the equation in standard form of a hyperbola with vertices (1,-3) and (-5,-3 and eccentricity 5/3
**
Equation is that 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.
x-coordinate of center=(1-5)/2=-2
y-coordinate of center=-3
center: (-2,-3)
..
length of horizontal transverse axis=6 (-5 to 1)=2a
a=3
a^2=9
..
eccentricity=c/a=c/3=5/3
c=5
c^2=25
..
c^2=a^2+b^2
b^2=c^2-a^2=25-9=16
b=4
..
Equation:
(x+2)^2/9-(y+3)^2/16=1
|
|
|
| |
