How do I graph the hyperbola represented by #4x^2-y^2-16x-2y+11=0#?

1 Answer
Sep 17, 2015

Answer:

Steps are explained below

Explanation:

Write the equation in the standard form:

#4(x^2-4x +2) - (y^2+2y+1) -8+1+11=0#

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

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

This hyperbola is therefor centered at (2, -1). It has a vertical transverse axis. Vertices are at (2,1) and (2,-3). From the center vertices would be 2 units up and down.

Now draw a rectangle centered at (2, -1) with length 4, along its transverse axis and width2 units. Draw both diagonals and extend on both sides. These are the asymptotes of the hyperbola and curve can be sketched from both vertices, along its asymptotes.

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