SOLUTION: Hello, I'm looking for some guidance with this question, can anyone help?
For the following equation of a hyperbola determine the center, vertices, foci, and asymptotes:
36x^2&
Algebra.Com
Question 968003: Hello, I'm looking for some guidance with this question, can anyone help?
For the following equation of a hyperbola determine the center, vertices, foci, and asymptotes:
36x^2−y^2−24x+6y−41=0
Found 2 solutions by stanbon, lwsshak3:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
For the following equation of a hyperbola determine the center, vertices, foci, and asymptotes:
36x^2−y^2−24x+6y−41=0
36(x^2-(24/36)x) - (y^2-6y) = 41
-----
Complete the square::
36(x^2 - (2/3)x + (1/3)^2) - (y^2-6y+9) = 41 + 36(1/9) - 9
------
36(x-(1/3)^2 - (y-3)^2 = 36
------
Rewrite:
(x-(1/3))^2 - (y-3)^2/36 = 1
a = 6 ; b = 1 ; c = sqrt(36+1) = sqrt(37)
-------------------------------------
center: (1/3, 3)
vertices:: (1/3,3-6);(1/3,3+6)
------------------
foci:: ((1/3),-3-sqrt(37)),((1/3),9+sqrt(37))
-----
asymptotes: I'll leave that to you
-----------------
Cheers,
Stan H.
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
For the following equation of a hyperbola determine the center, vertices, foci, and asymptotes:
36x^2−y^2−24x+6y−41=0
***
36x^2−24x−y^2+6y=41
36(x^2−(24/36)x)−(y^2-6y)=41
36(x^2−(2/3)x)−(y^2-6y)=41
complete the square:
36(x^2−(2/3)x+1/9)−(y^2-6y+9)=41+4-9
36(x-1/3)^2-(y-3)^2=36
given hyperbola has a horizontal transverse axis
Its standard form of equation:
For given hyperbola:
center: (1/3, 3)
a^2=1
a=1
vertices: (1/3±a,3)=(1/3±1,3)=(-2/3, 3) and (4/3, 3)
..
foci:
b^2=36
b=6
c^2=a^2+b^2=1+36=37
c=√37≈6.1
foci: (1/3±c,3)=(1/3±6.1,3)=(-5.8, 3) and (6.4, 3)
..
Asymptotes: (two straight-line equations(mx+b) that intersect at center
For hyperbolas with horizontal transfer axis, slopes of asymptotes=±b/a=±6/1=±6
..
Equation of asymptote with positive slope:
y=6x+b
solve for b using coordinates of center
3=6*1/3+b
b=1
equation: y=6x+1
..
Equation of asymptote with negative slope:
y=-6x+b
solve for b using coordinates of center
3=-6*1/3+b
b=5
equation: y=-6x+5
RELATED QUESTIONS
Hello :)
I'm struggling with this question and need some guidance to it. I can solve... (answered by Alan3354)
Hi, everyone. Can anyone help me with this question, please?
The area of a regular... (answered by Alan3354)
I'm looking for some needed assistance to help solve this equation by the addition... (answered by mananth)
i could use some help understanding this question.
in the equation of the hyperbola... (answered by venugopalramana)
hello there,
i'm looking for help solving this problem:
For the function f(x) =... (answered by dkppathak)
Hello, looking for some help with this problem.
5sqrt3(2sqrt15+sqrt8)
Thanks!
(answered by Theo,MathTherapy)
Hi everyone, Im looking for some help and guidance with the following graphing and... (answered by solver91311)
Hi everyone, Im looking for some help and guidance with the following graphing and... (answered by rfer)
Hello! My name is Laila. I need help with this question: Evaluate the following... (answered by rfer)