1. ## minimisation question

Suppose that $a\geq 0$. Find the least distance from the point [tex](a,0) to the ellipse $\frac{x^2}{4}+\frac{y^2}{1}=1$.

Well basically i used the distance formula, so $D=\sqrt{(x-a)^2+y^2}=\sqrt{(x-a)^2+1-\frac{x^2}{4}}$.

Then i set up a function $f(x)=(x-a)^2+1-\frac{x^2}{4}$.

I differentiated it it find $f'(x)=2(x-a)-\frac{x}{2}$.

I then found $x=\frac{4a}{3}$ to be the minimum turning point and subbed it back into distance formula to get $D=\sqrt{1-\frac{a^2}{3}}$.

least distance $=\begin{cases}
&\sqrt{1-\frac{a^2}{3}} \text{ if } 0\leq a\leq 3/2 \\
&\left | a-2 \right | \text{ if } a> 3/2
\end{cases}$

i get why its |a-2| if a>2 but i dont get why its |a-2| between a=3/2 and a=2.

2. Setting sqrt(1-a^2/3) = 2- a

we get a = 3/2 this is the point where the point distance from(a,0) to a point on the ellipse is the same as the distance from (a,0) to the vertex(2,0)
In fact for a = sqrt(3) sqrt(1-a^2/3) = 0 and sqrt(1-a^2/3) is imaginary

for a > sqrt(3)

If you want I made an animation demonstrating this on my web site:

Scratch Paper

3. hm...i see where you're getting at, but how come the distance formula stops working between a=3/2 and a=sqrt(3). It's still defined between these values. Also, if you let a=1.6 (say), then sqrt(1- (1.6^2 /3)) < |1.6-2| . Shouldnt this be the least distance in this case? And also, how come the distance formula gives 0 when a=sqrt(3). Surely, theres no point on the ellipse that gives this distance.

4. Note x= 4a/3 at the min pt

and y= sqrt(1-x^2/4)

if x = 4a/3 y = sqrt[1 - (4a^2)/9]

If a = 3/2 x = 2 and y = 0 you are at the vertex

if a > 3/2 you are no longer on the ellipse and the formula

D =sqrt(1-a^2/3) is no longer valid as you are not on the ellipse

you're right as sqrt(3) > 3/2 you don't have to worry about a point on the ellipse whose distance is 0

thats also why you don't have to worry about a= 1.6 either

5. thanks for clarifying that

6. From a=1.5 to 2, the shortest distance to the ellipse is 2-a, or a-2 if a>2.