Question

How do you find the standard form of `x^2 + 4y^2 - 8x + 16y + 16 =0` and what kind of a conic is it?

Answer 1

An ellipse with center `(4,-2)`, x-axis radius: `4`, and y-axis radius: `2`

Explanation:

Given:
`color(white)("XXX")x^2+4y^2-8x+16y+16=0`

Group `x` and `y` terms (and constant) separately as
`color(white)("XXX")color(blue)(x^2-8x)+color(red)(4(y^2+4y))+color(green)(16)=0`

Complete the squares for both the `x` and the `y` sub-expressions:
`color(white)("XXX")color(blue)(x^2-8x+16) + color(red)(4(y^2+4y+4))+color(green)(16)-color(blue)(16)-color(red)(4(4))=0`

Reduce `x` and `y` sub-expressions to squared binomials and move constant to the right side:
`color(white)("XXX")(x-4)^2+4(y+2)^2= 16`
or
`color(white)("XXX")(x-4)^2+4(y+2)^2= 4^2`

Divide both sides by `(4^2)`
and replace `(y+2)` with `(y-(-2))`
`color(white)("XXX")((x-4)^2)/(4^2)+(y-(-2))^2/(2^2)=1`

Note that the general form of an ellipse is
`color(white)("XXX")(x-h)^2/(a^2)+((y-k)^2)/(b^2)=1`
with
`color(white)("XXX")`center at `(h,k)`,
`color(white)("XXX")`x-axis radius: `(a)`, and
`color(white)("XXX")`y-axis radius: `(b)`
graph{x^2+4y^2-8x+16y+16=0 [-2.195, 10.29, -4.895, 1.345]}