|
Question 482730: Find the equation of a hyperbola whose asymptote coincide withe the points (-7, 4) and (-1, -14). Its other asymptote contains point (-4,1), and (-3/4, -53/4) is a point on the hyperbola.
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Find the equation of a hyperbola whose asymptote coincide with the points (-7, 4) and (-1, -14). Its other asymptote contains point (-4,1), and (-3/4, -53/4) is a point on the hyperbola.
**
I will assume the given point -53/4 was meant to be written as (-5,3/4); otherwise, it didn't seem to be correct.
..
Asymptotes are straight lines of the standard form: y=mx+b, m=slope, b=y-intercept
For asymptote with points (-7,4) and (-1,-14):
slope, m =∆y/∆x=(-14-4)/(-1,-(-7))=-18/6=-3
Equation:
y=-3x+b
solving for b using one of given points, (-7,4)
4=-3*-7+b
4=21+b
b=-17
Equation of asymptote:
y=-3x-17
..
For asymptote with points (-4,1) and (-3,4):
slope, m =∆y/∆x=(4-1)/(-3,-(-4)=3/1=3
Equation:
y=3x+b
solving for b using one of given points, (-4,1)
1=3*(-4)+b
1=-12+b
b=13
Equation of asymptote:
y=3x+13
..
Point of intersection of asymptotes=center of hyperbola
-3x-17=3x+13
6x=-30
x=-5
y=3x+13=-15+13=-2
Center of hyperbola: (-5,-2)
..
Note that the given point on the hyperbola(-5,3/4) is on the same line as the center (-5,2) which means the hyperbola has a vertical transverse axis of the standard form: (y-k)^/a^2-(x-h)^2/b^2=1.
The point (-5,3/4) are also the coordinates of the (upper) vertex.
..
a=half the length of the vertical transverse axis or the distance from the center to the vertex=2+3/4=11/4
a=11/4
slope =3=a/b
b=a/3=(11/4)/3=11/12
b=11/12
..
We now have the information needed to write the equation of the given hyperbola.
(y+2)^2/(11/4)^2-(x+5)^2/(11/12)^2=1
see graph below as a visual check on the answer
..
y=±(7.5625+7.5625(x+5)^2/.8403)^.5-2
|
|
|
| |
