Chapter 10.3, Problem 6GCE

To determine

Guess the behavior of the graph using given condition.

Explanation of Solution

Given:

The given equation is $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$ and the given condition is that the value of $x$ is replaced by $\left(x\pm c\right)$ .

Calculation:

The general equation of the ellipse is $\frac{{\left(x-h\right)}^2}{a^2}+\frac{{\left(y-k\right)}^2}{b^2}=1$ or $\frac{{\left(y-k\right)}^2}{a^2}+\frac{{\left(x-h\right)}^2}{b^2}=1$ .

If $a^2$ is the denominator of the $x$ term, the major axis is parallel to the x-axis. If $a^2$ is the denominator of the $y$ term, the major axis is parallel to the y-axis.

Here, only change in the value of $\text{'}h\text{'}$ by $c$ units.

So, the new graph shift $c$ units to the right (when the $x$ is replaced by $\left(x-c\right)$ and the new graph shift $c$ units to the left (when the $x$ is replaced by $\left(x+c\right)$ .

Example

When the $x$ is replaced by $\left(x-4\right)$ .

Original equation is

  $\frac{{\left(x-3\right)}^2}{18}+\frac{{\left(y+2\right)}^2}{32}=1$

In the equation, $a^2$ is the denominator of the $y$ term.

  $a^2=32$ , $b^2=18$ and $c^2=a^2-b^2=\sqrt{{32}^2-{18}^2}=10\sqrt{7}$

Center $\to \left(h,k\right)=\left(3,-2\right)$

Foci $\to \left(h,k\pm c\right)=\left(3,-2\pm 10\sqrt{7}\right)$

Major axis vertices $\to \left(h,k\pm a\right)=\left(3,-2\pm 4\sqrt{2}\right)$

Minor axis vertices $\to \left(h\pm b,k\right)=\left(3\pm 3\sqrt{2},-2\right)$

Substitute the value of $x$ with $\left(x-4\right)$ .

  $\begin{array}{l}\frac{{\left(x-4-3\right)}^2}{18}+\frac{{\left(y+2\right)}^2}{32}=1\\ \Rightarrow \frac{{\left(x-7\right)}^2}{18}+\frac{{\left(y+2\right)}^2}{32}=1\end{array}$

New equation is $\frac{{\left(x-7\right)}^2}{18}+\frac{{\left(y+2\right)}^2}{32}=1$ .

Here, only change in the value of $\text{'}h\text{'}$ by $4$ units.

So, the new graph shift $4$ units to the right.

Graph of the functions are given below.

  

In the graph, blue curve shows the original equation and the red curve shows the $4$ units shifted graph (to the right).

When the $x$ is replaced by $\left(x+4\right)$ .

Original equation is

  $\frac{{\left(x-3\right)}^2}{18}+\frac{{\left(y+2\right)}^2}{32}=1$

In the equation, $a^2$ is the denominator of the $y$ term.

  $a^2=32$ , $b^2=18$ and $c^2=a^2-b^2=\sqrt{{32}^2-{18}^2}=10\sqrt{7}$

Center $\to \left(h,k\right)=\left(3,-2\right)$

Foci $\to \left(h,k\pm c\right)=\left(3,-2\pm 10\sqrt{7}\right)$

Major axis vertices $\to \left(h,k\pm a\right)=\left(3,-2\pm 4\sqrt{2}\right)$

Minor axis vertices $\to \left(h\pm b,k\right)=\left(3\pm 3\sqrt{2},-2\right)$

Substitute the value of $x$ with $\left(x+4\right)$ .

  $\begin{array}{l}\frac{{\left(x+4-3\right)}^2}{18}+\frac{{\left(y+2\right)}^2}{32}=1\\ \Rightarrow \frac{{\left(x+1\right)}^2}{18}+\frac{{\left(y+2\right)}^2}{32}=1\end{array}$

New equation is $\frac{{\left(x+1\right)}^2}{18}+\frac{{\left(y+2\right)}^2}{32}=1$ .

Here, only change in the value of $\text{'}h\text{'}$ by $4$ units and the value of $h$ is negative.

So, the new graph shift $4$ units to the left.

Graph of the functions are given below.

  

In the graph, blue curve shows the original equation and the red curve shows the $4$ units shifted graph (to the left).