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Question 581789: pls help me.
1.asymptotes are y-x-1=0 and y+x-1=0 and y intercepts are 3 and -1. find the eq. of the hyperbola.
2.Focus at (0,25) asymptotes intersecting at (0,5) an asymptote passing through (24,13) Find the eq. of hyperbola
thank you. it will help a lot for me.
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! 1.asymptotes are y-x-1=0 and y+x-1=0 and y intercepts are 3 and -1. find the eq. of the hyperbola.
Standard form of equation for hyperbola with vertical transverse axis:
(y-k)^2/a^2-(x-h^2/b^2=1
given y-intercepts show that hyperbola has a vertical transverse axis.
y-coordinate of center=1 (halfway between 3 and -1 on the y-axis
x-coordinate of center=0
center: (0,1)
slope of asymptotes=±1 (from given equations of asymptotes)
y=x+1
y=-x+1
slope of asymptotes also=a/b=1
a=b
..
equation:
(y-k)^2/a^2-(x-h^2/b^2=1
(y-1)^2/a^2-x^2/b^2=1
using y-intercept point (0,3)
(3-1)^2/a^2-0=1
a^2=4
a=2=b
Equation of given hyperbola:
(y-1)^2/4-x^2/4=1
...
2.Focus at (0,25) asymptotes intersecting at (0,5) an asymptote passing through (24,13) Find the eq. of hyperbola.
**
Given focus (0,25) shows hyperbola has a vertical transverse axis
Asymptotes intersect at center
center: (0,5)
Using two points, (0,5) and (24,13) of one of the asymptotes:
slope of asymptote =∆y/∆x=(13/5)/(24/0)=8/24=1/3
slope also=a/b=1/3
b=3a
..
c=20 (from 5 to 25 on y-axis)
c^2=a^2+b^2
400=a^2+(3a)^2=10a^2
a^2=40
a=√40
..
b=3a=3√40
b^2=9*40=360
..
Equation of given hyperbola:
(y-5)^2/40-x^2/360=1
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