Question

How do you graph the ellipse `x^2/169+y^2/25=1` and find the center, the major and minor axis, vertices, foci and eccentricity?

Answer 1

Please see below.

Explanation:

This is the equation of an ellipse of the form `(x-h)^2/a^2+(y-k)^2/b^2=1`, whose center is `(h,k)`, major axis is `2a` and minor axis is `2b`. Vertices on major axis are `(h+-a,k)` and along minor axis are `(h,k+-b)`. Eccentricity is given by `e=sqrt(1-b^2/a^2)` and focii are `(h+-ae,k)`.

As `x^2/169+y^2/25=1` can be written as

`(x-0)^2/13^2+(y-0)^2/5^2=1`

Hence, this is an equation of an ellipse, whose center is `(0,0)`, major axis is `13xx2=26` and minor axis is `5xx2=10`.

Vertices are `(-13,0),(13,0),(5,0)` and `(-5,0)`.

Eccentricity is `e=sqrt(1-5^2/13^2)=sqrt(12^2/13^2)=12/13`

and fociie are `(+-(13xx12/13),0)` i.e. `(-12,0)` and `(12,0)`.

We can mark the four vertices, if so desired a few more points by using the equation `x^2/169+y^2/25=1`. For example, if `x=+-3`. `y=+-5sqrt(1-9/169)=+-20sqrt(10/169)=+-4.865` gives four more points `(3,4.865),(-3,4.865), (3,-4.865)` and `(-3,-4.865)`, joining which along with vertices will give us the desired ellipse.

The ellipse appears as shown below:

graph{(x^2/169+y^2/25-1)((x+13)^2+y^2-0.04)((x-13)^2+y^2-0.04)(x^2+(y+5)^2-0.04)(x^2+(y-5)^2-0.04)((x+12)^2+y^2-0.04)((x-12)^2+y^2-0.04)=0 [-20, 20, -10, 10]}