Chapter 11.3, Problem 24E

To determine

To find: the equation for the hyperbola.

Answer to Problem 24E

The equation of hyperbola is $\frac{x^2}{12}-\frac{y^2}{64}=1$ .

Explanation of Solution

Given information:

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

Vertices of hyperbola : $\left(\pm 2\sqrt{3},0\right)$

pointon hyperbola : $\left(4,4\right)$

Calculation:

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

So the equation is of the form,

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

Since $a=2\sqrt{3}$ and a point on the hyperbola , so to find the value of , substitute $x,y\text{ and }a$ value in the equation,

  $\begin{array}{l}\frac{x^2}{{\left(a\right)}^2}-\frac{y^2}{b^2}=1\\ \frac{{\left(4\right)}^2}{{\left(2\sqrt{3}\right)}^2}-\frac{{\left(4\right)}^2}{{\left(b\right)}^2}=1\text{ [substitute]}\\ \frac{16}{12}-\frac{16}{{\left(b\right)}^2}=1\text{ [square]}\\ \frac{16}{{\left(b\right)}^2}=\frac{16}{12}-1\text{ [simplify]}\\ \frac{16}{{\left(b\right)}^2}=\frac{4}{16}\text{ [subtract]}\\ {\left(b\right)}^2=64\text{ [simplify]}\\ b=8\text{ [sqaure root]}\end{array}$

so the equation of hyperbola is,

  $\begin{array}{l}\frac{x^2}{{\left(a\right)}^2}-\frac{y^2}{b^2}=1\\ \frac{x^2}{{\left(2\sqrt{3}\right)}^2}-\frac{y^2}{{\left(8\right)}^2}=1\text{ [substitute]}\\ \frac{x^2}{12}-\frac{y^2}{64}=1\end{array}$

Hence, the equation of hyperbola is $\frac{x^2}{12}-\frac{y^2}{64}=1$ .