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?

3 Answers
Mar 11, 2018

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

Mar 11, 2018

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

Mar 12, 2018

(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