Question

How do you put `x^2-2x+2y^2-12y+3=0` in standard form, find the center, the endpoints, vertices, the foci and eccentricity?

Answer 1

Given:
`x^2-2x+2y^2-12y+3=0`

Now,
`x^2-2x=x^2-2x+1-1`
`=(x-1)^2-1`
`2y^2-12y=2(y^2-6y)`
`(y^2-6y)=y^2-6y+9-9`
`=(y-3)^2-9`
`2y^2-12y=2((y-3)^2-9)`

The given equation becomes
`(x-1)^2-1+2((y-3)^2-9)+3=0`

Simplifying
`(x-1)^2-1+2(y-3)^2-18+3=0`
`(x-1)^2+2(y-3)^2=18+1-3`
`(x-1)^2+2(y-3)^2=16`

Dividing throughout by 16, we have
`(x-1)^2/16+2(y-3)^2/16=16/16`
`(x-1)^2/4^2+(y-3)^2/(16/2)=1`
`(x-1)^2/4^2+(y-3)^2/(2sqrt(2))^2=1`

Here,
the major axis is `a=4`
the minor axis is `b=2sqrt(2)`

Centre is `(1,3)`

End points:
`(5,3)`
`1,3+2sqrt(2)`
`(-3,3)`
`(1,3-2sqrt(2))`

Eccentricity is
`e=sqrt((a^2-b^2)/a^2)`
`a^2=16`
`b^2=8`
`sqrt((a^2-b^2)/b^2)=sqrt((16-8)/16)`
`=sqrt(8/16)`

`e=0.707`