# How do I find the points on the ellipse 4x^2 + y^2 = 4 that are furthest from (1, 0)?

May 27, 2016

The furthest points are located at { (x = -0.333333, y = -1.88562), (x = -0.333333, y = 1.88562) :}

#### Explanation:

The ideas explained below are based in the hypothesis that the involved functions are twice continuous.

The problem can be stated as: given a set of points $\left(x , y\right)$ defined through the relationship $f \left(x , y\right) = 4 {x}^{2} + {y}^{2} - 4 = 0$, and a fixed point $q = \left({x}_{q} , {y}_{q}\right)$, determine a point ${p}_{o} = \left({x}_{o} , {y}_{o}\right)$ in the set defined by $f \left(x , y\right) = 0$ such that the distance $d \left(x , y , x\right) = \left\lVert {p}_{o} - q \right\rVert$ has maximum value.

If such point exists, then the normal to $f \left(x , y\right) = 0$ in ${p}_{o}$ must be linearly dependent with the normal to $d \left(x , y , z\right)$ in ${p}_{o}$. In other words, the two normals must be co-lineal. Formally this condition is equivalent to

$\nabla f \left({x}_{o} , {y}_{o}\right) = \lambda \nabla d \left({x}_{o} , {y}_{o}\right)$
where $\nabla = \left(\frac{\partial}{\partial x} , \frac{\partial}{\partial x}\right)$

or

$\left\{\begin{matrix}{f}_{x} = \lambda {d}_{x} \\ {f}_{y} = \lambda {d}_{y}\end{matrix}\right.$

or

$\left\{\begin{matrix}8 x = \lambda \frac{x - {x}_{q}}{d \left(x y\right)} \\ 2 y = \lambda \frac{y - {y}_{q}}{d \left(x y\right)}\end{matrix}\right.$

or

$\left\{\begin{matrix}8 x = {\lambda}_{d} \left(x - {x}_{q}\right) \\ 2 y = {\lambda}_{d} \left(y - {y}_{q}\right)\end{matrix}\right.$
here ${\lambda}_{d} = \frac{\lambda}{d \left(x , y\right)}$
Putting all together we have the conditions:

$\left\{\begin{matrix}4 {x}_{o}^{2} + {y}_{o}^{2} - 4 = 0 \\ 8 {x}_{o} = {\lambda}_{d} \left({x}_{o} - {x}_{q}\right) \\ 2 {y}_{o} = {\lambda}_{d} \left({y}_{o} - {y}_{q}\right)\end{matrix}\right.$

Now making
${p}_{q} = \left({x}_{q} , {y}_{q}\right) = \left(1 , 0\right)$ as proposed
and solving for ${x}_{o} , {y}_{o} , {\lambda}_{d}$ we have

{ (x = -1., y = 0., lambda_d = 4.), (x = -0.333333, y = -1.88562, lambda_d = 2.), (x = -0.333333, y = 1.88562, lambda_d= 2.) :}

The computed distances are respectively

$\left(2. , 2.3094 , 2.3094\right)$