Chapter 11.3, Problem 39E

To determine

To find: the equation for the hyperbola.

Answer to Problem 39E

The equation of hyperbola is $\frac{x^2}{16}-\frac{y^2}{16}=1$ .

Explanation of Solution

Given information:

From the figure, provided in the question it is observed that

Point on hyperbola : $\left(5,3\right)$

asymptotes of hyperbola : $y=\pm x$

Calculation:

let the hyperbola has a horizontal transverse axis.

So the equation is of the form,

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

For transverse axis hyperbola, the asymptote

  $y=\pm \frac{b}{a}x$

So from the $y=\pm x$ asymptote it can be observed that ,

  $\begin{array}{l}\frac{b}{a}=\frac{1}{1}\\ a=b\end{array}$

Since $a=b$ and a point on the hyperbola , so to find the value of , substitute $x,y\text{ and }a$ value in the equation,

  $\begin{array}{l}\frac{y^2}{{\left(a\right)}^2}-\frac{x^2}{b^2}=1\\ \frac{{\left(5\right)}^2}{{\left(b\right)}^2}-\frac{{\left(3\right)}^2}{{\left(b\right)}^2}=1\text{ [substitute]}\\ \frac{25}{{\left(b\right)}^2}-\frac{9}{{\left(b\right)}^2}=1\text{ [square]}\\ \frac{16}{{\left(b\right)}^2}=1\text{ [simplify]}\\ {\left(b\right)}^2=\frac{16}{1}\text{ [cross multiply]}\\ {\left(b\right)}^2=16\\ b=4\text{ [square root]}\end{array}$

So to find value of $a$ ,

  $\begin{array}{l}a=b\\ a=4\text{ [substitute]}\end{array}$

so the equation of hyperbola is,

  $\begin{array}{l}\frac{x^2}{{\left(a\right)}^2}-\frac{y^2}{b^2}=1\\ \frac{x^2}{{\left(4\right)}^2}-\frac{y^2}{{\left(4\right)}^2}=1\text{ [substitute]}\\ \frac{x^2}{16}-\frac{y^2}{16}=1\text{ [simplify] }\end{array}$

Hence, the equation of hyperbola is $\frac{x^2}{16}-\frac{y^2}{16}=1$ .