Chapter 9.3, Problem 45E

To determine

To find:The standard form of the equation of hyperbola when the vertices are $\left(1,2\right)$

  $\left(3,2\right)$ and asymptotes are $y=x$ , $y=4-x$ .

Answer to Problem 45E

The standard form of the equation of hyperbola is $\frac{{\left(x-2\right)}^2}{1}-\frac{{\left(y-2\right)}^2}{1}=1$ .

Explanation of Solution

Given information:

Thevertices of hyperbola are $\left(1,2\right)$

  $\left(3,2\right)$ and asymptotes of hyperbola is $y=x$ , $y=4-x$ .

Calculation:

The vertices of the hyperbola is given by $\left(h+a,k\right)$ and $\left(h-a,k\right)$ . Since the given vertices are $\left(1,2\right)$ and $\left(3,2\right)$ .The center of the hyperbola is the midpoint of the line joining the two vertices.

So the co-ordinates of the centre are

  $\begin{array}{l}\left(\frac{3+1}{2},\frac{2+2}{2}\right)\\ \left(2,2\right)\end{array}$

The centre of the hyperbola is given by $\left(h,k\right)$ . So by comparing it we get, $h=2$ and $k=2$ .

The asymptotes of the hyperbola is given by $y=\pm \frac{b}{a}\left(x-h\right)+k$ . The asymptotes of the hyperbola given is $y=x$ , $y=4-x$ .So by comparing it with the equation $y=x$ , $y=4-x$ we get

  $a=1$ , $b=1$ .

The standard equation of the hyperbola is given as $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$ .

Substitute $a=1$ , $b=1$ , $h=2$ and $k=2$ to find the standard equation of the hyperbola

  $\begin{array}{l}\frac{{\left(x-2\right)}^2}{1^2}-\frac{{\left(y-2\right)}^2}{1^2}=1\\ {\left(x-2\right)}^2-{\left(y-2\right)}^2=1\end{array}$

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