How do you write a standard form equation for the hyperbola with #y^2 - 4y - 4x^2 + 8x = 4#?

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Answer 1 · 2018-07-16T12:40:16

The equation is #(y-2)^2/4-(x-1)^2/1=1#

The general equation for a hyperbola is

#(x-h)^2/a^2-(y-k)^2/b^2=1#

or

#(y-k)^2/a^2-(x-h)^2/b^2=1#

Here, we have

#y^2-4y-4x^2+8x=4#

Complete the square

#(y^2-4y+4)-4(x^2-2x+1)=4+4-4#

#(y-2)^2-4(x-1)^2=4#

Dividing by #4#

#(y-2)^2/4-4(x-1)^2/4=1#

#(y-2)^2/4-(x-1)^2/1=1#

The center is #C=(1,2)# and the vertices are #A=(1,0)# and #A'=(1,4)#

graph{((y-2)^2/4-(x-1)^2/1-1)=0 [-20.43, 20.11, -9, 11.28]}

Answer 2 · 2018-07-16T12:56:27

Standard form of vertical hyperbola equation is
#(y-2)^2/2^2 -(x-1)^2/(1^2) = 1#

# y^2-4 y -4 x^2+8 x= 4# or

# (y^2-4 y +4) -4(x^2-2 x +1) = 4-4+4# or

# (y-2)^2 -4(x-1)^2 = 4# or

#(y-2)^2/4 -(cancel4(x-1)^2)/cancel4 = 1# or

#(y-2)^2/2^2 -(x-1)^2/(1^2) = 1#

The standard form of equation of vertical hyperbola is

#(y-k)^2/a^2-(x-h)^2/b^2=1#

Therefore, the standard form of vertical hyperbola equation

is #(y-2)^2/2^2 -(x-1)^2/(1^2) = 1#

graph{y^2-4 y -4 x^2+ 8 x -4=0 [-40, 40, -20, 20]} [Ans]