Chapter 10.5, Problem 39PPSE

Solution

To determine: The equation by choosing two points on the same axis to be the vertices of a hyperbola and similarly choose two another points of same axis to the foci.

The required equations is $\frac{x^2}{4^2}-\frac{y^2}{3^2}=1$ .

Given Information:

Chose two points on the same axis to be the vertices of a hyperbola and similarly choose two another points of same axis to the foci.

Calculation:

Consider the given information,

The vertices of the hyperbola is defined for horizontal as,

  $\left(h\pm a,k\right)$

And, the foci is defined as,

  $\left(h\pm c,k\right)$

Suppose that the point of vertices is $\left(\pm 4,0\right)$ and the foci point is $\left(\pm 5,0\right)$ .

Rewrite both the points following,

  $\left(\pm 4,0\right)=\left(0\pm 4,0\right)$

And,

  $\left(\pm 5,0\right)=\left(0\pm 5,0\right)$

Compare with the vertices and foci points.

  $\begin{array}{l}a=4\\ c=5\end{array}$

Use the formula $c^2=a^2+b^2$ and calculate the value of b .

  $\begin{array}{c}{5}^2=4^2+b^2\\ b^2=25-16\\ =9\end{array}$

Substitute all the values in general equation of hyperbola $\frac{{\left(x-h\right)}^2}{a^2}-\frac{{\left(y-k\right)}^2}{b^2}=1$ .

  $\begin{array}{c}\frac{{\left(x-0\right)}^2}{{\left(4\right)}^2}-\frac{{\left(y-0\right)}^2}{9}=1\\ \frac{x^2}{16}-\frac{y^2}{9}=1\end{array}$

Therefore, the obtained equation is $\frac{x^2}{4^2}-\frac{y^2}{3^2}=1$ .