Question

How do you graph the ellipse `(x+5)^2/16+(y-4)^2/1=1` and find the center, the major and minor axis, vertices, foci and eccentricity?

Answer 1

Please see below.

Explanation:

`(x-h)^2/a^2+(y-k)^2/b^2=1` is the equation of an ellipse,

whose center is `(h,k)`, `2a` and `2b` are major axis and minor axis (if `a>b`, but if `a< b`, it is reverse). In the former case major axis is parallel to `x`-axis and in latter case major axis is parallel to minor axis.

vertices are `(h,k+-b)` and `(h+-a,k)`

Eccentricity `e` can be obtained using `e=sqrt(1-("minor axis")^2/("major axis")^2)`

and focii are at a distance of `ae` on either side of center along major axis (or `be` if major axis is parallel to `y`-axis).

Hence, in `(x+5)^2/16+(y-4)^2/1=1`

center is `(-5,4)`, major axis is `2sqrt16=8`, which is parallel to `x`-axis and minor axis is `2sqrt1=2`

Vertices are `(-5,4+-1)` and `(-5+-4,4)` i.e. `(-5,5)`, `(-5,3)`, `(-1,4)` and `(-9,4)`.

Eccentricity is `e=sqrt(1-1/4)=sqrt3/2` and `ae=sqrt3`

and focii are `(-5+-sqrt3,4)` i.e. `(-5-sqrt3,4)`and `(-5+sqrt3,4)`.

They appear as follows:

graph{((x+5)^2+16(y-4)^2-16)((x+5)^2+(y-4)^2-0.01)((x+5)^2+(y-5)^2-0.01)((x+5)^2+(y-3)^2-0.01)((x+1)^2+(y-4)^2-0.01)((x+9)^2+(y-4)^2-0.01)((x+5+sqrt3)^2+(y-4)^2-0.01)((x+5-sqrt3)^2+(y-4)^2-0.01)=0 [-10.12, -0.12, 1.46, 6.46]}