SOLUTION: Find the foci of a hyperbola with the equation 9y2 - 72y - 16x2 - 64x - 64 = 0.
Algebra.Com
Question 145227: Find the foci of a hyperbola with the equation 9y2 - 72y - 16x2 - 64x - 64 = 0.
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Find the foci of a hyperbola with the equation 9y2 - 72y - 16x2 - 64x - 64 = 0.
-------------------
9(y^2 - 8y + 16) - 16(x^2 -4x + 4) = 64+9*16-4*16
---------
9(y -4)^2 - 16(x - 2)^2 = 144
-------
[(y -4)^2/16] - [(x - 2)^2/4] = 1
---------
Center at (2,4); a = 4 ; b = 2 ; therefore c = sqrt(a^2+b^2) = 2sqrt(5)
Focus at (2, 4+2sqrt(5)) and (2,4-2sqrt(5))
=============
Cheers,
Stan H.
RELATED QUESTIONS
i need a lot of help with these questions i do not understand them at all..! pleas help.! (answered by lwsshak3)
More hard ones.
1. What is the midpoint of the segment joining (3, 5) and (-6, 11)?
(answered by longjonsilver)
Find the vertices and foci of the hyperbola.
16x2 − y2 − 64x − 4y + 44 (answered by lwsshak3)
Convert the equation to standard form. Locate the foci and find the equation of the... (answered by Alan3354,greenestamps)
Find the foci for the given hyperbola.
9y2-x2-36y-2x+26=0
y2=ysquared... (answered by venugopalramana)
Write an equation of a hyperbola with the given foci and vertices.
foci (+-17, 0),... (answered by tommyt3rd)
Write an equation of a hyperbola with the given foci and vertices
Foci (0,+-13),... (answered by MathLover1)
A conic has an equation of an asymptote equal to 3x=4y. What is the equation of the conic (answered by greenestamps,MathLover1)
Find the standard form of the equation of the hyperbola with the given characteristics
(answered by lwsshak3)