Chapter 9, Problem 4CT

To determine

To find: Find the center, vertices, foci and asymptotes of hyperbola represented by $x^2-4y^2-4x=0$ .

Answer to Problem 4CT

Center: $(2,0)$ ; Foci: $\left(2+\sqrt{5},0\right)\text{ and }\left(2-\sqrt{5},0\right)$ ; Vertices: $(4,0)\text{ and }(0,0)$ and

Asymptotes: $y=\frac{x}{2}-1\text{ and }y=-\frac{x}{2}+1$ .

Explanation of Solution

Given information: Hyperbola: $x^2-4y^2-4x=0$

Calculations:

  $\begin{array}{l}{x}^2-4y^2-4x=0\\ \Rightarrow x^2-4x-4y^2=0\\ \Rightarrow x^2-2(x)(2)+2^2-2^2-4y^2=0\left(\text{Add and subtract }{2}^2\right)\\ \Rightarrow {\left(x-2\right)}^2-4-4y^2=0\left[\because {\text{ a}}^2-2ab+b^2={\left(a-b\right)}^2\right]\\ \Rightarrow {\left(x-2\right)}^2-4y^2=4\\ \Rightarrow \frac{{\left(x-2\right)}^2}{4}-\frac{4y^2}{4}=1\\ \Rightarrow \frac{{\left(x-12\right)}^2}{4}-\frac{y^2}{1}=1\end{array}$

So, $\frac{{\left(x-2\right)}^2}{4}-\frac{y^2}{1}=1$ is a standard form of equation of hyperbola.

Comparing this equation to standard equation of hyperbola: $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$

  $\begin{array}{l}(h,k)\equiv (2,0)\\ \Rightarrow h=2,k=0\\ a^2=4\\ \Rightarrow a=\sqrt{4}\\ \Rightarrow a=2\\ b^2=1\\ \Rightarrow b=1\\ c=\sqrt{a^2+b^2}\\ \Rightarrow c=\sqrt{4+1}\\ \Rightarrow c=\sqrt{5}\end{array}$

Since, $a>b$ .

So, major axis of hyperbola is parallel to x-axis and minor axis is parallel to y-axis.

Center: $(h,k)\equiv (2,0)$

  $\begin{array}{l}h+c=2+\sqrt{5}\\ h-c=2-\sqrt{5}\end{array}$

Foci: $(h+c,k)\equiv \left(2+\sqrt{5},0\right)\text{ and }(h-c,k)\equiv \left(2-\sqrt{5},0\right)$

  $\begin{array}{l}h+a=2+2\\ \Rightarrow h+a=4\\ h-a=2-2\\ \Rightarrow h-a=0\end{array}$

Vertices: $(h+a,k)\equiv (4\text{,0) and }(h-a,k)\equiv (0,0)$

  $\begin{array}{l}\text{Asymptotes: }y=\frac{b(x-h)}{a}+k\text{ and }y=-\frac{b(x-h)}{a}+k\\ \Rightarrow y=\frac{(1)(x-2)}{2}+0\text{ and }y=-\frac{(1)(x-2)}{2}+0\\ \Rightarrow y=\frac{x}{2}-\frac{2}{2}\text{ and }y=-\frac{x}{2}+\frac{2}{2}\\ \Rightarrow y=\frac{x}{2}-1\text{ and }y=-\frac{x}{2}+1\end{array}$

Hence, Center: $(2,0)$ ; Foci: $\left(2+\sqrt{5},0\right)\text{ and }\left(2-\sqrt{5},0\right)$ ; Vertices: $(4,0)\text{ and }(0,0)$ and

Asymptotes: $y=\frac{x}{2}-1\text{ and }y=-\frac{x}{2}+1$ .

Graph: