Question

How do you find the horizontal and vertical tangents of `4x^2 + y^2 -8x +4y +4=0`?

Answer 1

Horizontal tangents at: `(1,0)` and `(1,-4)`
Vertical tangents at: `(0,-2)` and `(2,-2)`

Explanation:

We have:

`4x^2 + y^2 -8x +4y +4=0`

Differentiating implicitly:

`8x + 2ydy/dx -8 +4dy/dx = 0`
`:. (y+2)dy/dx =4-4x`
`:. dy/dx =(4-4x)/(y+2)`

We have horizontal tangents when `dy/dx=0`

`=> 4-4x = 0`
`:. \ \ x = 1`

With `x=1` we have:

`4 + y^2 -8 +4y +4=0`
`:. y^2 +4y =0`
`:. y(y+4) =0`
`:. y=0,-4`

ie the coordinates `(1,0)` and `(1,-4)`

We have vertical tangents when `dy/dx rarr oo` (Note I did not write `dy/dx = oo` as a mathematician I really dislike this abuse of notation, `oo` is not a number, and cannot be equated).

`=> y+2 = 0`
`:. \ \ y = -2`

With `y=-2` we have:

`4x^2 + 4 -8x -8 +4=0`
`:. 4x^2 -8x = 0`
`:. 4x(x-2) = 0`
`:. x=0,2`

ie the coordinates `(0,-2)` and `(2,-2)`

We can confirm this by completing the square:

`4x^2 -8x +y^2 + 4y +4=0`
`4{x^2 -2x} +{y^2 + 4y} +4=0`
`4{(x-1)^2-1} +{(y+2)^2 - 4} +4=0`
`4(x-1)^2 -4+(y+2)^2 -4+4=0`
`4(x-1)^2 +(y+2)^2 =4`
`(x-1)^2 +(y+2)^2/4 =1`
`(x-1)^2 +((y+2)/2)^2 =1`

An ellipse of centre `(1,-2)` with minor and major axis `1` and `2` respectively.

graph{4x^2 + y^2 -8x +4y +4=0 [-1, 3, -6, 2]}