How do I graph the hyperbola with the equation $4 x^{2} - y^{2} + 4 y - 20 = 0 ?$ $4x^2−y^2+4y−20=0?$ ?

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Answer 1 · 2018-08-14T12:41:16

Please see the explanation below

You need the center, the vertices and the asymptotes.

The equation is

$4 x^{2} - y^{2} + 4 y - 20 = 0$ $4x^2-y^2+4y-20=0$

$4 x^{2} - (y^{2} - 4 y + 4) = 20 - 4 = 16$ $4x^2-(y^2-4y+4)=20-4=16$

Dividing by $16$ $16$

$\frac{4}{16} x^{2} - \frac{{(y - 2)}^{2}}{16} = 1$ $4/16x^2-(y-2)^2/16=1$

$\frac{{(x)}^{2}}{4} - \frac{{(y - 2)}^{2}}{16} = 1$ $(x)^2/4-(y-2)^2/16=1$

Comparing this to the general equation of a hyperbola

$\frac{{(x - h)}^{2}}{a^{2}} - \frac{{(y - k)}^{2}}{b^{2}} = 1$ $(x-h)^2/a^2-(y-k)^2/b^2=1$

The center is $C = (h, k) = (0, 2)$ $C=(h,k)=(0,2)$

The vertices are

$A = (h + a, k) = (0 + 2, 2) = (2, 2)$ $A=(h+a,k)=(0+2,2)=(2,2)$

and

$A'=(h-a,k)=(0-2,2)=(-2,2)$

And the asymptotes are

$(y-2)^2/16=(x)^2/4$

$y-2=+-(2x)$

$y-2=2x$ and $y-2=-2x$

Now, you can plot the graph

graph{(4x^2-y^2+4y-20)(y-2-2x)(y-2+2x)=0 [-15.77, 16.27, -5.8, 10.2]}

How do I graph the hyperbola with the equation 4x^2−y^2+4y−20=0?4x2−y2+4y−20=0?4x^2−y^2+4y−20=0??