Question

Convert `x^2+2y^2-2x+8y-11=0` to standard form of equation for ellipse and find its vertices, focii and latus rectum?

Answer 1

Standard form of ellipse is `(x-1)^2/(2sqrt5)^2+(y+2)^2/(sqrt10)^2=11`. Vertices along major axis are `(1-2sqrt5,-2)` and `(1+2sqrt5,-2)` and those along minor axis are `(1,-2-sqrt10)` and `(1,-2+sqrt10)`. Focii are `(1-sqrt10,-2)` and `(1+sqrt10,-2)` and latus rectum is `2sqrt5`.

Explanation:

Let us convert this to the form of a conic.

`x^2+2y^2-2x+8y-11=0` can be written as

`(x^2-2x+1)+2(y^2+4y+4)-11-8-1=0`

or `(x-1)^2+2(y+2)^2=20`

i.e. `(x-1)^2/20+(y+2)^2/10=1`

or `(x-1)^2/(2sqrt5)^2+(y+2)^2/(sqrt10)^2=11`, which is standard form.

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)`. Latus rectum is `(2b^2)/a`

Hence, here center is `(1,-2)`, major axis is `4sqrt5` and minor axis is `2sqrt10`.

Vertices along major axis are `(1-2sqrt5,-2)` and `(1+2sqrt5,-2)` and those along minor axis are `(1,-2-sqrt10)` and `(1,-2+sqrt10)`.

Eccentricity is `sqrt(1-10/20)=1/sqrt2` and `ae=2sqrt5/sqrt2=sqrt10`

Hence, focii are `(1-sqrt10,-2)` and `(1+sqrt10,-2)` and latus rectum is `(2xx10)/sqrt20=sqrt20=2sqrt5`

graph{(x^2+2y^2-2x+8y-11)((x-1+2sqrt5)^2+(y+2)^2-0.02)((x-1-2sqrt5)^2+(y+2)^2-0.02)((x-1)^2+(y+2-sqrt10)^2-0.02)((x-1)^2+(y+2+sqrt10)^2-0.02)((x-1+sqrt10)^2+(y+2)^2-0.02)((x-1-sqrt10)^2+(y+2)^2-0.02)=0 [-9.29, 10.71, -7.68, 2.32]}