SOLUTION: how can i find the focus/foci and vertex/vertices x^2-4y^2-2x+16y=20 this is how far i have gotten, but i am not sure if it is right so far x^2-4y^2-2x+16y=20 (x^2-2x+1)-

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: how can i find the focus/foci and vertex/vertices x^2-4y^2-2x+16y=20 this is how far i have gotten, but i am not sure if it is right so far x^2-4y^2-2x+16y=20 (x^2-2x+1)-      Log On


   

Question 706005: how can i find the focus/foci and vertex/vertices
x^2-4y^2-2x+16y=20
this is how far i have gotten, but i am not sure if it is right so far
x^2-4y^2-2x+16y=20
(x^2-2x+1)-4(y^2+4y+4)=20+1+4
(x-1)^2/25-4(y+4)^2/25=25/25
(x-1)^2/25-4(y+4)^2/25=1
please help. thank you.
Found 2 solutions by solver91311, lwsshak3:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!




Note the -4 coefficient OUTSIDE the parentheses so the 1st degree y term has to have a coefficient of -4 (-4 times -4 is +16), then in order to complete the square inside the parentheses you need a +4, and +4 times -4 is -16. What you add to the left must be added to the right, hence the -16 in the RHS.
From there:





So and

Can you take it from there?

John

Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it
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Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
how can i find the focus/foci and vertex/vertices
x^2-4y^2-2x+16y=20
complete the square:
x^2-2x-4y^2+16y=20
(x^2-2x+1)-4(y^2-4y+4)=20+1-16
(x-1)^2-4(y-2)^2=5
%28x-1%29%5E2%2F5-4%28y-2%29%5E2%2F%285%2F4%29=1
This is an equation of a hyperbola with horizontal transverse axis.
Its standard form: %28x-h%29%5E2%2Fa%5E2-%28y-k%29%5E2%2Fb%5E2=1, (h,k)=(x,y) coordinates of the center.
For given equation:
Center: (1,2)
a=5
a^2=25
vertices: (1±a,2)=(1±5,2)=(-4,2) and (6,2)
..
b=5/4
b^2=25/16
..
foci
c^2=a^2+b^2=25+(25/16)=425/16
c=√425/4≈5.15
foci:(1±c,2)=(1±5.15,2)=(-4.15,2) and (6.15,2)