# How do you graph the ellipse (x-2)^2/4+(y+3)^2/9=1 and find the center, the major and minor axis, vertices, foci and eccentricity?

Explanation given below

#### Explanation:

The given equation of ellipse:

$\setminus \frac{{\left(x - 2\right)}^{2}}{4} + \setminus \frac{{\left(y + 3\right)}^{2}}{9} = 1$

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

Above equation of ellipse is in standard form ${\left(x - h\right)}^{2} / {a}^{2} + {\left(y - k\right)}^{2} / {b}^{2} = 1 \setminus \quad \left(\setminus \forall \setminus a < b\right)$

The center of ellipse : $\left(h , k\right) \setminus \equiv \left(2 , - 3\right)$

Major axis: $2 b = 2 \left(3\right) = 6$

Minor axis: $2 a = 2 \left(2\right) = 4$

The vertices : $\left(h \setminus \pm a , k\right)$ & $\left(h , k \setminus \pm b\right)$

$\left(2 \setminus \pm 2 , - 3\right)$ & $\left(2 , - 3 \setminus \pm 3\right)$

Eccentricity $e = \setminus \sqrt{1 - {a}^{2} / {b}^{2}} = \setminus \sqrt{1 - {2}^{2} / {3}^{2}} = \setminus \frac{\sqrt{5}}{3}$

Focii : $\left(h , k \setminus \pm b e\right)$

$\left(2 , - 3 \setminus \pm 3 \setminus \cdot \setminus \frac{\sqrt{5}}{3}\right) \setminus \equiv \left(2 , - 3 \setminus \pm \setminus \sqrt{5}\right)$

The equations of axes of ellipse: $y + 3 = 0$ & $x - 2 = 0$

Plot the axes of ellipse. Specify the center $\left(2 , - 3\right)$, vertices & focii on the corresponding axes. Draw a free hand curve passing through the vertices & the points of intersection of ellipse with coordinate axes to get graph.