# Thread: Stuck on this problem

1. ## Stuck on this problem

Find the points on the ellipse 4x^2+y^2=4 that are farthest away from the point (1,0).

2. Originally Posted by DroMan
Find the points on the ellipse 4x^2+y^2=4 that are farthest away from the point (1,0).
$\displaystyle 4x^2 + y^2 = 4$
$\displaystyle y = \pm \sqrt4-4x^2$

find the distance

$\displaystyle D: \sqrt (x-1)^2 + (y-1)^2$
$\displaystyle \sqrt -3x^2 -2x + 5$

take the derivative and set it to 0.

$\displaystyle dd/dx = \frac{1}{2} (3x^2-2x+5) ^ (-1/2)(-6x-2)$
$\displaystyle \frac{-6x-2}{2 \sqrt -3x^2 - 2x + 5} = 0$

multiply by the denom.

$\displaystyle -6x-2=0$

$\displaystyle -6=2$

$\displaystyle x= \frac {-1}{3}$

now plug into the ellipse and solve for y.

3. Let $\displaystyle d$ be the distance from the point (1,0) to the ellipse.

Now if you remember your distance formula: $\displaystyle d = \sqrt{(x - x_1)^2 + (y - y_1)^2} \ \Leftrightarrow \ d^2 = (x - x_1)^2 + (y - y_1)^2$

Here, $\displaystyle (x_1, y_1) = (1,0)$ and $\displaystyle (x , y )$ is any arbitrary point on the ellipse.

Now, arranging your equation of your ellipse we get: $\displaystyle y^2 = 4-4x^2$

So, we want to find the absolute maximum of: $\displaystyle d^2 = (x-1)^2 + y^2 \ \Leftrightarrow \ {\color{blue}d^2 = (x-1)^2 + (4-4x^2)}$

You do this by taking the derivative and setting it equal to 0 (note that taking the derivative of $\displaystyle d^2$ is equal to $\displaystyle 2d d'$ by chain rule). Then determine if it is a maximum and you'll be done.

4. Originally Posted by euclid2
$\displaystyle 4x^2 + y^2 = 4$
$\displaystyle y = \pm \sqrt4-4x^2$

find the distance

$\displaystyle D: \sqrt (x-1)^2 + (y-1)^2$
$\displaystyle \sqrt -3x^2 -2x + 5$

take the derivative and set it to 0.

$\displaystyle dd/dx = \frac{1}{2} (3x^2-2x+5) ^ (-1/2)(-6x-2)$
$\displaystyle \frac{-6x-2}{2 \sqrt -3x^2 - 2x + 5} = 0$

multiply by the denom.

$\displaystyle -6x-2=0$

$\displaystyle -6=2$

$\displaystyle x= \frac {-1}{3}$

now plug into the ellipse and solve for y.
how did you get the points for the distance formula? i do not understand what you did after, "multiply by the denom."

5. Originally Posted by DroMan
how did you get the points for the distance formula? i do not understand what you did after, "multiply by the denom."
Let me be more specific with the answer.
So now d=√(√(4-4x²)-0)²+(x-1)²) for the first half of the ellipse. simplify
d=√(4-4x²+x²-2x+1)
=√(-3x²-2x+5)
Now find derivative
$\displaystyle d'=(1/2)(-3x²-2x+5)^(-1/2)(-6x-2)$
set equal to zero
$\displaystyle (1/2)(-3x²-2x+5)^(-1/2)(-6x-2)$
which means
$\displaystyle -6x-2=0$
$\displaystyle x=2/-6=-1/3$
Now we want maximum distance so we must test to see if this is a maximum.
$\displaystyle d'(-1)=1$ <==positive
$\displaystyle d'(0)=-5^(-1/2)$ <==negative
Therefore a maximum occurs at x=-1/3
So now we find y value on ellipse
$\displaystyle 4x²+y²=4$
$\displaystyle 4(-1/3)²+y²=4$
$\displaystyle y²=32/9$
y=±√(32/9)=±(4√2)/3
Now even though we defined the ellipse in two parts it turns out that there's no need to attempt to find a maximum using the second half of the ellipse. Why? Take a look at this.
The second half of the ellipse was defined asy2=-√(4-4x²)
So a point on the second half of the ellipse is defined as (x, -√(4-4x²))
The distance between a point on the second half of the ellipse and the point (1, 0) is d=√((-√(4-4x²)-0)²+(x-1)²)= √(4-4x²+x²-2x+1)]. Notice that this is the same thing we started with above? So we'd be optimizing the same thing again.
Anyways, maximum distance is between[tex] (-1/3, (-4√2)/3) and (1, 0)and from(-1/3, (4√2)/3) to (1, 0)