Chapter 10, Problem 6CT

To determine

The equation of the hyperbola and graph of the equation with center $(2,2)$, vertex $(2,4)$ and contains the point $\left(2+\sqrt{10},5\right)$

Answer to Problem 6CT

Solution:

The equation of the hyperbola with center $(2,2)$, vertex $(2,4)$ and contains the point $\left(2+\sqrt{10},5\right)$ is $\frac{{(y-2)}^2}{4}-\frac{{\left(x-2\right)}^2}{8}=1$ and the graph of the equation is

  

Explanation of Solution

Given information:

The hyperbola has center $(2,2)$, vertex $(2,4)$ and contains the point $\left(2+\sqrt{10},5\right)$

The center $(2,2)$ and vertex $(2,4)$ both lie on the vertical line $x=2$

Thus, the transverse axis is parallel to the $y-$ axis.

The distance from the center $(2,2)$ to a vertex $(2,4)$ is $a=2$, so the other vertex must be $(2,0)$

The general form of the equation of hyperbola is $\frac{{(y-k)}^2}{a^2}-\frac{{(x-h)}^2}{b^2}=1$

Substitute $h=2,k=2$ and $a=2$ ,

  $\Rightarrow \frac{{(y-2)}^2}{2^2}-\frac{{(x-2)}^2}{b^2}=1$

  $\Rightarrow \frac{{(y-2)}^2}{4}-\frac{{(x-2)}^2}{b^2}=1$

Since the graph contains the point $(x,y)=\left(2+\sqrt{10},5\right)$, it satisfies the equation of hyperbola.

  $\Rightarrow \frac{{(5-2)}^2}{4}-\frac{{\left(2+\sqrt{10}-2\right)}^2}{b^2}=1$

  $\Rightarrow \frac{3^2}{4}-\frac{{\left(\sqrt{10}\right)}^2}{b^2}=1$

  $\Rightarrow \frac{9}{4}-\frac{10}{b^2}=1$

  $\Rightarrow \frac{10}{b^2}=\frac{9}{4}-1$

  $\Rightarrow b^2=8$

  $\Rightarrow b=2\sqrt{2}$

Therefore, the equation of the hyperbola is $\frac{{(y-2)}^2}{4}-\frac{{\left(x-2\right)}^2}{8}=1$ .

Now, $c^2=a^2+b^2=4+8=12$

  $\Rightarrow c=2\sqrt{3}$

The distance from the center to either focus is $c=2\sqrt{3}$ . Thus, the foci are $F_1=(h,k+c)=\left(2,2+2\sqrt{3}\right)$ and $F_2=(h,k-c)=\left(2,2-2\sqrt{3}\right)$ .

The asymptotes are given by the lines $y-k=\pm \frac{a}{b}(x-h)$ .

Thus, the asymptotes are $y=\frac{\sqrt{2}}{2}(x-2)+2$ and $y=-\frac{\sqrt{2}}{2}(x-2)+2$ .

To graph the equation of hyperbola, form a rectangle using the vertices $\left(2,4\right),\left(2,0\right)$ and the points $\left(h\pm b,k\right)$ : $\left(2-2\sqrt{2},2\right),\left(2+2\sqrt{2},2\right)$

By using all the above information, the graph of the hyperbola $\frac{{(y-2)}^2}{4}-\frac{{\left(x-2\right)}^2}{8}=1$ is as follows: