Chapter 11.3, Problem 23E

To determine

To find: the equation for the hyperbola.

Answer to Problem 23E

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

Explanation of Solution

Given information:

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

Vertices of hyperbola : $\left(0,\pm 4\right)$

pointon hyperbola : $\left(3,-5\right)$

Calculation:

since from the given vertices it is observed that the vertices of the hyperbola are on the y-axis, so the hyperbola has a vertical transverse axis.

So the equation is of the form,

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

Since $a=4$ and a point on the hyperbola , so to find the value of , substitute $x,y\text{ and }a$ value in the equation,

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

so the equation of hyperbola is,

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

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