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

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

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Answer 1 · 2018-03-11T20:37:54

The maximum is $\frac{y}{x} = 2 \frac{\sqrt{3}}{3}$ $y/x = 2sqrt3/3$

Explanation:

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

$y = 2 \sqrt{1 - {(x - 2)}^{2}}$ $y = 2sqrt(1-(x-2)^2)$

Substitute into $\frac{y}{x}$ $y/x$

$2 \frac{\sqrt{1 - {(x - 2)}^{2}}}{x}$ $2sqrt(1-(x-2)^2)/x$

Differentiate with respect to x:

$\frac{d (2 \frac{\sqrt{1 - {(x - 2)}^{2}}}{x})}{d x} = \frac{6 - 4 x}{x^{2} \sqrt{1 - {(x - 2)}^{2}}}$ $(d(2sqrt(1-(x-2)^2)/x))/dx = (6-4x)/(x^2sqrt(1-(x-2)^2))$

Set the first derivative equal to 0:

$\frac{6 - 4 x}{x^{2} \sqrt{1 - {(x - 2)}^{2}}} = 0$ $(6-4x)/(x^2sqrt(1-(x-2)^2)) = 0$

$x = \frac{3}{2}$ $x = 3/2$

$y = 2 \sqrt{1 - {(\frac{3}{2} - 2)}^{2}}$ $y = 2sqrt(1-(3/2-2)^2)$

$y = \sqrt{3}$ $y = sqrt3$

$\frac{y}{x} = 2 \frac{\sqrt{3}}{3}$ $y/x = 2sqrt3/3$

Answer 2 · 2018-03-11T20:44:41

See below.

Explanation:

Making $\frac{y}{x} = λ$ $y/x = lambda$ and substituting into $4 {(x - 2)}^{2} + y^{2} = 4$ $4(x-2)^2+y^2=4$ we have

$4 {(x - 2)}^{2} + λ^{2} x^{2} = 4$ $4(x-2)^2+lambda^2x^2 = 4$ and then

$λ^{2} = \frac{4 (1 - {(x - 2)}^{2})}{x^{2}}$ $lambda^2 = (4(1-(x-2)^2))/x^2$

The maximum for $λ$ $lambda$ is the same for $λ^{2}$ $lambda^2$

The maximum condition is

$\frac{d}{d λ} (λ^{2}) = - (\frac{8 (2 x - 3)}{x^{3}}) = 0$ $d/(d lambda) (lambda^2) = -((8 (2 x-3))/x^3)=0$

then

$x_{0} = \frac{3}{2}$ $x_0 = 3/2$ with $y_{0} = \pm 2 \sqrt{1 - {(x_{0} - 2)}_{0}^{2}} = \pm \sqrt{3}$ $y_0 = pm2sqrt(1-(x_0-2)^2)= pm sqrt3$

are the solution points and then $λ = \frac{2}{3} \sqrt{3}$ $lambda =2/3 sqrt3$

Answer 3 · 2018-03-12T10:37:09

$\frac{2 \sqrt{3}}{3}$ $(2 sqrt3)/3$

Explanation:

Considering

$4 {(x - 2)}^{2} + y^{2} = 4 \Rightarrow {(x - 2)}^{2} + \frac{1}{4} y^{2} = 1$ $4(x-2)^2+y^2=4 rArr (x-2)^2+1/4 y^2=1$

now making the change of variables

$u = x - 2$ $u = x-2$
$v = \frac{y}{2}$ $v=y/2$ we get

${(x - 2)}^{2} + \frac{1}{4} y^{2} = 1 \equiv u^{2} + v^{2} = 1$ $(x-2)^2+1/4 y^2=1 equiv u^2+v^2=1$

now the condition

$\frac{y}{x} = λ \equiv \frac{2 v}{u + 2} = λ \Rightarrow - \frac{1}{2} u + \frac{1}{λ} v = 1$ $y/x=lambda equiv (2v)/(u+2) = lambda rArr -1/2 u+1/lambda v=1$

so the problem in the new coordinates is.

Determine $λ$ $lambda$ such that the line

$- \frac{1}{2} u + \frac{1}{λ} v = 1$ $-1/2 u+1/lambda v=1$

is tangent to the circle

$u^{2} + v^{2} = 1$ $u^2+v^2 = 1$

Now given a point $p_{0}$ $p_0$ pertaining to the circumference we have

$p = (u, v)$ $p = (u,v)$
$p_{0} = ({cos θ}_{0}, {sin θ}_{0})$ $p_0 = (cos theta_0,sin theta_0)$
$\to n = ({cos θ}_{0}, {sin θ}_{0})$ $vec n = (cos theta_0, sin theta_0)$

and by construction, the line

$L \to p = p_{0} + λ \to n$ $L->p = p_0 + lambda vec n$ or the non-parametric version

$u {cos θ}_{0} + v {sin θ}_{0} = 1$ $u cos theta_0 + v sin theta_0 = 1$

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

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

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

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