Chapter 11.2, Problem 3E

The graph of the equation $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$ with a > b > 0 is an ellipse with vertices (_____, _____ ) and (_____, _____) and foci (0, ± c), where c = __________. So the graph of $\frac{x^2}{4^2}+\frac{y^2}{5^2}=1$is an ellipse with vertices (_____, _____) and (_____, _____) and foci (_____, _____ ) and (_____, _____).

To determine

The graph of the equation $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$ with $a>b>0$ is an ellipse with vertices $(0,\pm a)$, $(\pm b,0)$ and foci $(0,\pm c)$, where $c=\sqrt{a^2-b^2}$. So the graph of $\frac{x^2}{4^2}+\frac{y^2}{5^2}=1$ is an ellipse with vertices $(0,\pm 5)$ and $(\pm 4,0)$ and foci are $(0,-3)$ and $(0,+3)$.

Explanation of Solution

Definition used:

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

Calculation:

By the definition of the ellipse stated above conclude that, vertices of the ellipse $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$ are $(0,\pm a)$, $\left(\pm b,0\right)$ and the foci are $F(0,\pm c)$. Also, $c=\sqrt{a^2-b^2}$.

Compare the equation $\frac{x^2}{4^2}+\frac{y^2}{5^2}=1$ with the ellipse $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$,

$a^2=5^2$

$a=5$(Take square root on both sides)

And,

$b^2=4^2$

$b=4$(Take square root on both sides)

Clearly, $a=5$, $b=4$ and $a>b>0$. Hence, by the above definition the vertices are $(0,\pm 5)$ and $(\pm 4,0)$.

Also,

$c=\sqrt{a^2-b^2}$

$c=\sqrt{5^2-4^2}$

$c=\sqrt{25-16}$

$c=\sqrt{9}$

$c=\pm 3$.

Hence, by the above definition of ellipse the foci are $(0,-3)$ and $(0,+3)$.

Therefore, the vertices and the foci of the ellipse  $\frac{x^2}{4^2}+\frac{y^2}{5^2}=1$ are $(0,\pm 5)$, $(\pm 4,0)$ and $(0,-3)$ and $(0,+3)$, respectively.