Question

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?

Answer 1

Explanation given below

Explanation:

The given equation of ellipse:

`\frac{(x-2)^2}{4}+\frac{(y+3)^2}{9}=1`

`\frac{(x-2)^2}{2^2}+\frac{(y+3)^2}{3^2}=1`

Above equation of ellipse is in standard form `(x-h)^2/a^2+(y-k)^2/b^2=1\quad (\forall \ a < b)`

The center of ellipse : `(h, k)\equiv (2, -3)`

Major axis: `2b=2(3)=6`

Minor axis: `2a=2(2)=4`

The vertices : `(h\pma, k)` & `(h, k\pmb)`

`(2\pm2, -3)` & `(2, -3\pm3)`

Eccentricity `e=\sqrt{1-a^2/b^2}=\sqrt{1-2^2/3^2}=\sqrt5/3`

Focii : `(h, k\pm be)`

`(2, -3\pm 3\cdot \sqrt5/3)\equiv(2, -3\pm\sqrt5)`

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

Plot the axes of ellipse. Specify the center `(2, -3)`, 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.