How do I graph the hyperbola with the equation #4x^2−y^2−16x−2y+11=0=0#?

1 Answer
Sep 18, 2015

Answer:

As explained below

Explanation:

Convert this equation in the standard form of a vertical hyperbola as, #((y+1)^2)/4 - ((x-2)^2)/1=1#

This Hyperbola as its center at (2,-1) and its vertices at (2,1) and (2,-3) on its vertical transverse axis.

Draw its asymptotes y+1 =2(x-2) and y+1=-2(x-2). The curve can be sketched now as shown below.

Alternatively, draw a rectangle of length 4 units and width 2 units with its center at (2,-1). Extend its diagonals on both sides, which would be the asymptotes of the hyperbola. Sketch the curve as shown below.

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