Question

Point `P(x, y)` is on the ellipse with equation `4(x-2)^2+y^2=4`. Find the largest possible value of `y/x`?

Answer 1

The maximum is `y/x = 2sqrt3/3`

Explanation:

Solve `4(x-2)^2+y^2=4` for the postive value of y:

`y = 2sqrt(1-(x-2)^2)`

Substitute into `y/x`

`2sqrt(1-(x-2)^2)/x`

Differentiate with respect to x:

`(d(2sqrt(1-(x-2)^2)/x))/dx = (6-4x)/(x^2sqrt(1-(x-2)^2))`

Set the first derivative equal to 0:

`(6-4x)/(x^2sqrt(1-(x-2)^2)) = 0`

`x = 3/2`

`y = 2sqrt(1-(3/2-2)^2)`

`y = sqrt3`

`y/x = 2sqrt3/3`

Answer 2

See below.

Explanation:

Making `y/x = lambda` and substituting into `4(x-2)^2+y^2=4` we have

`4(x-2)^2+lambda^2x^2 = 4` and then

`lambda^2 = (4(1-(x-2)^2))/x^2`

The maximum for `lambda` is the same for `lambda^2`

The maximum condition is

`d/(d lambda) (lambda^2) = -((8 (2 x-3))/x^3)=0`

then

`x_0 = 3/2` with `y_0 = pm2sqrt(1-(x_0-2)^2)= pm sqrt3`

are the solution points and then `lambda =2/3 sqrt3`

Answer 3

`(2 sqrt3)/3`

Explanation:

Considering

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

now making the change of variables

`u = x-2`
`v=y/2` we get

`(x-2)^2+1/4 y^2=1 equiv u^2+v^2=1`

now the condition

`y/x=lambda equiv (2v)/(u+2) = lambda rArr -1/2 u+1/lambda v=1`

so the problem in the new coordinates is.

Determine `lambda` such that the line

`-1/2 u+1/lambda v=1`

is tangent to the circle

`u^2+v^2 = 1`

Now given a point `p_0` pertaining to the circumference we have

`p = (u,v)`
`p_0 = (cos theta_0,sin theta_0)`
`vec n = (cos theta_0, sin theta_0)`

and by construction, the line

`L->p = p_0 + lambda vec n` or the non-parametric version

`u cos theta_0 + v sin theta_0 = 1`

is tangent to the circle.

Now comparing

`{(-1/2 u+1/lambda v=1),(u cos theta_0 + v sin theta_0 = 1):}`

`cos theta_0 = -1/2 rArr theta_0 = (2pi)/3`

hence

`lambda = 1/(sin theta_0) =1/sin((2pi)/3) = (2 sqrt3)/3`