Question

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 $(0,\pm c)$ wherec=…………... So the graph of $\frac{x^2}{4^2}+\frac{y^2}{5^2}=1$is an ellipse with vertices (…… , ……..)and foci (……. , …….).

Short answer

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

Step-by-step solution

Step 1. Given Information.

Finding the vertices $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$ of the elipse

The vertices are (0,a) and (0,-a).

Step 2. Finding the value of c.

The value of c is given by:

$\begin{array}{c}{c}^2=a^2-b^2.\\ \Rightarrow c=\sqrt{a^2-b^2}.\end{array}$

Given ellipse $\frac{x^2}{4^2}+\frac{y^2}{5^2}=1$comparing with $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$, we get

$\begin{array}{l}a=5,\\ b=4\text{and}\\ c=\sqrt{a^2-b^2}\\ \Rightarrow c=\sqrt{5^2-4^2}\\ \Rightarrow c=\pm 3\end{array}$

Therefore,the vertices are (0,5) and (-o,5)

The foci are:

$\begin{array}{l}(0,c)=(0,3)\text{and}\\ (0,-c)=(0,-3)\end{array}$