Chapter 3.1, Problem 13CFU

To determine

To find: the coordinates of three additional points on the graph and other x-intercept.

Answer to Problem 13CFU

Three additional points on the graph are $\left(-6,\frac{3\sqrt{11}}{5}\right),\left(6,-\frac{3\sqrt{11}}{5}\right)$ and $\left(-6,-\frac{3\sqrt{11}}{5}\right)$ .

The other x-intercept is $\left(5,0\right)$ .

Explanation of Solution

Given information:

The equation is as follows:

  $\frac{x^2}{25}-\frac{y^2}{9}=1$

Points on graph equation are $\left(6,\frac{3\sqrt{11}}{5}\right)$ .

Point on x-intercept is $-5$ .

Calculation:

Put $\left(a,b\right)$ into the equation and solve as follows:

  $\begin{array}{l}\frac{x^2}{25}-\frac{y^2}{9}=1\\ \frac{a^2}{25}-\frac{b^2}{9}=1\end{array}$

If the solution of equation is same as that of any other case, then their graphs are symmetric.

Let’s check whether the given cases are symmetric with the given equation or not by substituting replacing $\left(x,y\right)$with different coordinates in the equation and simplify as shown below.

Check symmetry with	$\left(x,y\right)$	$\frac{x^2}{25}-\frac{y^2}{9}=1$	Symmetric with graph equation (Yes/No)
x-axis	$\left(a,-b\right)$	$\begin{array}{c}\frac{a^2}{25}-\frac{{\left(-b\right)}^2}{9}=1\\ \frac{a^2}{25}-\frac{b^2}{9}=1\end{array}$	Yes
y-axis	$\left(-a,b\right)$	$\begin{array}{c}\frac{{\left(-a\right)}^2}{25}-\frac{b^2}{9}=1\\ \frac{a^2}{25}-\frac{b^2}{9}=1\end{array}$	Yes

Since the graph is symmetric to y-axis, so the another x-intercept is $\left(5,0\right)$ . Also any value of x can give the same value of y because of the symmetry with y-axis. Thus, one of the point coordinate can be $\left(-6,\frac{3\sqrt{11}}{5}\right)$ .

Since the graph is symmetric to x-axis also, so any value of y can give the same value of x. Thus, other two points can be $\left(-6,-\frac{3\sqrt{11}}{5}\right)$ and $\left(6,-\frac{3\sqrt{11}}{5}\right)$ .