SOLUTION: How do I identify the equation (name the conic section), and then convert the equation to its standard form by completing the square on x and y
for the equation
4y^2-9x^2-16y-36x
Algebra.Com
Question 391623: How do I identify the equation (name the conic section), and then convert the equation to its standard form by completing the square on x and y
for the equation
4y^2-9x^2-16y-36x-56=0
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
given:4y^2-9x^2-16y-36x-56=0
4y^2-16y-9x^2-36x -56 =0
4(y^2-4y+4)-9(x^2-4x+4) =56+16-36
4(y-2)^2-9(x+2)^2=36
÷36
(y-2)^2/9 - (x+2)^2/4 =1
This is a hyperbola with its center at (-2,2), and transverse axis vertical.
standard form of hyperbola, (y-k)^2/a^2-(x-h)^2 =1
a^2 =9
a=3
b^2=4
b=2
c=sqrt(a^2)+(b^2
sqrt(13)
length of transverse axis = 2a =2*3=6
length of conjugate axis=2b=2*2=4
vertices=(-2,2±a)=(2,2±3)
foci =(-2,2±c)=(2±,sqrt(13)
asymptotes = (3/2)x+5 and (-3/2)x-1
see graph below
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