Graphing Ellipses An equation of an ellipse is given. (a) Find the vertices, foci, and eccentricity of the ellipse.

(b) Determine the lengths of the major and minor axes. (c) Sketch a graph of the ellipse.

$\frac{x^2}{49}+\frac{y^2}{25}=1$

The equation of the ellipse, $\frac{x^2}{49}+\frac{y^2}{25}=1$

If we compare the given equation $\frac{x^2}{49}+\frac{y^2}{25}=1$ with standard form

$\frac{{(x-h)}^2}{a^2}+\frac{{(y-k)}^2}{b^2}=1$.

This gives us

$\begin{array}{l}h=0,\\ k=0\\ a^2=49,\\ \Rightarrow a=7\\ b^2=25.\\ \Rightarrow b=5\end{array}$

So vertices of the ellipse are $(\pm a,0) or (\pm 7,0)$

Foci are \[(\pm c,0)\],where $c^2=a^2-b^2$

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

By substituting values we will get $c^2=49-25$

Therefore Foci are $(\pm 2\sqrt{6},0)$

Eccentricity $e=\frac{c}{a}.$ We have values of both, by substituting themwe get $e=\frac{2\sqrt{6}}{7}$

The equation of the ellipse, $\frac{x^2}{49}+\frac{y^2}{25}=1$

The equation of the given ellipse is $\frac{x^2}{49}+\frac{y^2}{25}=1$

By comparing the given equation with the general form we found that

$\begin{array}{l}a=7\\ b=5\end{array}$

Length of Major axis =2a

$\begin{array}{l}=2\left(7\right)\\ =14\end{array}$

Length of Minor axis=2b

$\begin{array}{l}=2\left(5\right)\\ =10\end{array}$

The equation of the given ellipse is $\frac{x^2}{49}+\frac{y^2}{25}=1$.

From the equation of ellipse, we can see that denominator of $x^2$ is greater than that of $y^2$. So this Would be a Horizontal ellipse. By comparing the given equation with the standard form equation $\frac{{(x-h)}^2}{a^2}+\frac{{(y-k)}^2}{b^2}=1$

$\frac{x^2}{49}+\frac{y^2}{25}=1$

We will get:

$\begin{array}{l}h=0\\ k=0\\ a^2=49\\ a=7\\ b^2=25\\ b=5\\ c=\sqrt{a^2-b^2}\\ c=\sqrt{24}\\ c=2\sqrt{6}\end{array}$

Foci of the ellipse $(\pm c,0)$ are $(\pm 2\sqrt{6},0)$, vertices $(\pm a,0)$ are $(\pm 7,0)$.

The graph will be