Question 611774: Write the equation of the hyperbola with foci at (1, 5) and (7, 5) and with vertices at (2, 5) and (6, 5).
My incorrect answer was (x-4)^2/4 - (y+5)^2/5 = 1
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Write the equation of the hyperbola with foci at (1, 5) and (7, 5) and with vertices at (2, 5) and (6, 5)
**
This is an equation of a hyperbola with horizontal transverse axis.
Its standard form of equation: (x-h)^2/a^2-(y-k)^2/b^2=1, (h,k)=(x,y) coordinates of center
x-coordinate of center=4 (midpoint between (1 and 7) or (2 and 6)
y-coordinate of center=5
center:(4,5)
length of horizontal transverse axis=4( from 2 to 6)=2a
a=2
a^2=4
2c=6 (from1 to 7)
c=3
c^2=9
c^2=a^2+b^2
b^2=c^2-a^2=9-4=5
Equation of given hyperbola:
(x-4)^2/4-(y-5)^2/5=1
you just missed the sign in the y-term
|
|
|
