Suppose that . Find the least distance from the point [tex](a,0) to the ellipse .
Well basically i used the distance formula, so .
Then i set up a function .
I differentiated it it find .
I then found to be the minimum turning point and subbed it back into distance formula to get .
However the answer is:
i get why its |a-2| if a>2 but i dont get why its |a-2| between a=3/2 and a=2.
Can someone please explain why?
Feb 5th 2010, 02:01 AM
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:
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.
Feb 5th 2010, 02:33 AM
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D =sqrt(1-a^2/3) is no longer valid as you are not on the ellipse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
Feb 5th 2010, 02:41 AM
thanks for clarifying that :)
Feb 5th 2010, 02:46 AM
From a=1.5 to 2, the shortest distance to the ellipse is 2-a, or a-2 if a>2.