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

Sep 18, 2015

As explained below

#### Explanation:

Convert this equation in the standard form of a vertical hyperbola as, $\frac{{\left(y + 1\right)}^{2}}{4} - \frac{{\left(x - 2\right)}^{2}}{1} = 1$

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.