Question 1142763: Calculus Question
Find the points on the ellipse:
4x^2 + y^2 = 4
that are farthest from (1, 0).
Hint: Begin with the distance formula.
Answer by greenestamps(13215)   (Show Source):
You can put this solution on YOUR website!
The ellipse has center (0,0); the semi-major axis is 2 units in the y direction; the semi-minor axis is 1 unit in the x direction.
Here is a graph showing the ellipse as two functions,  and 
By symmetry, it is clear that there will be two points that are farthest from (1,0); both with the same x coordinate, and with y coordinates that are opposites. So we only need to find one of the two points.
We could use calculus to maximize the distance between (1,0) and a point on the ellipse. However, the calculus is easier if we maximize the square of that distance; the distance will be maximum when the square of the distance is maximum.
Let D be the square of the distance between (1,0) and a point on the ellipse in the second quadrant.
Find where the derivative is zero:
ANSWER: The two points on the ellipse farthest from (1,0) are (-1/3,(4/3)*sqrt(2)) and (-1/3,-(4/3)*sqrt(2)).
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