Find the points on the ellipse 4x^2+y^2=4 that are farthest away from the point (1,0).
Let be the distance from the point (1,0) to the ellipse.
Now if you remember your distance formula:
Here, and is any arbitrary point on the ellipse.
Now, arranging your equation of your ellipse we get:
So, we want to find the absolute maximum of:
You do this by taking the derivative and setting it equal to 0 (note that taking the derivative of is equal to by chain rule). Then determine if it is a maximum and you'll be done.
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
set equal to zero
which means
Now we want maximum distance so we must test to see if this is a maximum.
<==positive
<==negative
Therefore a maximum occurs at x=-1/3
So now we find y value on ellipse
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)