Chapter 11.4, Problem 3E

To determine

To label: The vertices and foci on the ellipses $\frac{x^2}{5^2}+\frac{y^2}{4^2}=1$ and $\frac{{\left(x-3\right)}^2}{5^2}+\frac{{\left(y-1\right)}^2}{4^2}=1$

Answer to Problem 3E

For the ellipse $\frac{x^2}{5^2}+\frac{y^2}{4^2}=1$, the vertices are $V_1\left(-5,0\right)$ and $V_2\left(5,0\right)$, and the foci are $F_1\left(-3,0\right)$ and $F_2\left(3,0\right)$.

For the ellipse $\frac{{\left(x-3\right)}^2}{5^2}+\frac{{\left(y-1\right)}^2}{4^2}=1$, the vertices are $V_1\left(-2,1\right)$ and $V_2\left(8,1\right)$, and the foci are $F_1\left(0,1\right)$ and $F_2\left(6,1\right)$.

Explanation of Solution

Definition used:

Definition 1:

The equation of the ellipse with center at the origin, vertices $(\pm a,0)$ and foci $F(\pm c,0)$ is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, where $c=\sqrt{a^2-b^2}$, $a>b>0$.

Definition 2

If $h$ and $k$ are any two positive real numbers and a graph of any equation is in terms of $x$ and $y$, then

(a) When replace $x$ by $x-h$, the graph shifts right by $h$ units.

(b) When replace $x$ by $x+h$, the graph shifts left by $h$ units.

(c) When replace $y$ by $y-k$, the graph shifts upward by $k$ units.

(d) When replace $y$ by $y+k$, the graph shifts downward by $k$ units.

Calculations:

Consider the equation of ellipse with the center at $\left(0,0\right)$.

$\frac{x^2}{5^2}+\frac{y^2}{4^2}=1$

Consider the equation of the shifted ellipse with the center at $\left(3,1\right)$.

$\frac{{\left(x-3\right)}^2}{5^2}+\frac{{\left(y-1\right)}^2}{4^2}=1$ .

That is, equation $\left(2\right)$ may obtain from equation $\left(1\right)$ replacing $x$ by $x-3$ and $y$ by $y-1$.

Compare the equation $\left(1\right)$ with the general equation of ellipse, $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.

Then, $a^2=5^2$ implies $a=\pm 5$ and $b^2=4^2$ implies $b=\pm 4$.

Use the equation $c^2=a^2-b^2$ from definition 1 to find the value of c.

$\begin{array}{c}{c}^2=a^2-b^2\\ =25-16\left(\because a^2=25\text{ and }{b}^2=16\right)\\ c^2=9\\ c=\pm 3\end{array}$

Therefore, for the equation $\left(1\right)$, the vertices are $V_1\left(-5,0\right)$ and $V_2\left(5,0\right)$, and the foci are $F_1\left(-3,0\right)$ and $F_2\left(3,0\right)$.

For the ellipse $\frac{{\left(x-3\right)}^2}{5^2}+\frac{{\left(y-1\right)}^2}{4^2}=1$, the center is $C\left(3,1\right)$.

The terms $x-3$ and $y-1$ shifts the center of the ellipse, $\frac{x^2}{5^2}+\frac{y^2}{4^2}=1$, from the origin to the point, that is 3 units right and 1 unit upward.

Hence, the whole ellipse shifts, so the vertices and the foci.

Obtain the vertices as follows.

The sums $-5+3=-2\text{ and 5+3=8 }$ result in the x-coordinates of the vertices $V_1$ and $V_2$ of the ellipse, respectively and $0+1=1$ results in the y-coordinate of the vertices $V_1$ and $V_2$ of the ellipse, respectively.

Thus, $V_1\left(-2,1\right)$ and $V_2\left(8,1\right)$.

Similarly, obtain the foci as follows.

The sums $-3+3=0\text{ and 3+3=6 }$ result in the x-coordinates of the foci $F_1$ and $F_2$ of the ellipse, respectively and $0+1=1$ results in the y-coordinate of the foci $F_1$ and $F_2$ of the ellipse, respectively.

Thus, $F_1\left(0,1\right)$ and $F_2\left(6,1\right)$.

Therefore, for the equation $\left(2\right)$, the vertices are $V_1\left(-2,1\right)$ and $V_2\left(8,1\right)$, and the foci are $F_1\left(0,1\right)$ and $F_2\left(6,1\right)$.