Chapter 10, Problem 7RE

In Problems 1-10, identify each equation. If it is a parabola, give its vertex, focus, and directrix; if it is an ellipse, give its center, vertices, and foci: if it is a hyperbola, give its center, vertices, foci, and asymptotes.

$y^2-4y-\text{ 4}{x}^2\text{ + 8}x\text{ = }4$

To determine

Each equation. If it is a parabola, give its vertex, focus, and directrix; if it is an ellipse, give its center, vertices, and foci; if it is a hyperbola, give its center, vertices, foci, and asymptotes.

Answer to Problem 7RE

Hyperbola

Center: $\left(1,2\right)$

Vertices: $\left(1,4\right),\left(1,0\right)$

Foci: $\left(1,2\pm \surd 5\right)$

Asymptote: $\left(y-2\right)=\pm 2\left(x-1\right)$.

Explanation of Solution

Given:

$y²-4y-4x²+8x=4$

Formula used:

Equation	Center	Transverse axis	Foci	Vertices	Asymptotes
$\frac{{(x-h)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$ $b²=c²-a²$	$\left(h,k\right)$	$x\text{-axis}$	$\left(h+c,0\right)$	$\left(h+a,0\right)$	$y=\pm \frac{b}{a}x$
	$\left(h,k\right)$	$x\text{-axis}$	$\left(h-c,0\right)$	$\left(h-a,0\right)$

Calculation:

The equation $y²-4y-4x²+8x=4$ can be rewritten as,

$\left(y-2\right)²-4\left(x-1\right)²=4-4+4$

${\left(y-2\right)}^2-4{\left(x-1\right)}^2=4$

$\frac{{\left(x-1\right)}^2}{4}-\frac{{\left(y-2\right)}^2}{1}=1$

$\frac{{\left(x-1\right)}^2}{4}-\frac{{\left(y-2\right)}^2}{1}=1$

is of the form $\frac{{(x-h)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$

It represents a hyperbola.

$a=2,b=1,h=1,k=2$

Center at $\left(h,k\right)=\left(1,2\right)$.

Vertices: $\left(h,k\pm a\right)=\left(1,2\pm 2\right)=\left(1,4\right),\left(1,0\right)$.

$b^2=c^2-a^2$

$c=\sqrt{1+4}=\surd 5$

Foci: $\left(h,k\pm c\right)=\left(1,2\pm \surd 5\right)$.

Asymptotes: $\left(y-2\right)=\pm 2\left(x-1\right)$.