Question

The equation is 4x²+8y²+8x-4=8. How to find its vertices, co-vertices and it's Foci?

Answer 1

Please see the explanation below

Explanation:

Let's rearrange the equation

`4x^2+8y^2+8x-4=8`

`4x^2+8y^2+8x=8+4=12`

Dividing by `4`

`x^2+2y^2+2x=3`

Completing the square

`x^2+2x+2y^2=3`

`x^2+2x+1+2y^2=3+1`

`(x+1)^2+2y^2=4`

Dividing by `4`

`(x+1)^2/4+y^2/(4/2)=1`

`(x+1)^2/(2)^2+y^2/(sqrt(2))^2=1`

The general equation of an ellipse is

`(x-h)^2/a^2+(y-k)^2/b^2=1`

This is an equation of an ellipse, center `C=(h,k)=(-1,0)`

The vertices are `A=(h+a,k)=(-1+2,0)=(1,0)`

`A'=(h-a,k)=((-1)-2, 0)=(-3,0)`

`B=(h, k+b)=(-1, 0+sqrt2)=(-1, sqrt2)`

`B'=(h, k-b)=(-1, 0sqrt2)=(-1, -sqrt2)`

Calculate `c`

`c^2=a^2-b^2=4-2=2`

The foci are

`F=(h+c,k)=(-1+sqrt2, 0)`

And

`F'=(h-c, k), (-1-sqrt2,0)`

graph{x^2+2y^2+2x-3=0 [-5.308, 2.487, -1.806, 2.09]}