Question 1142763
<br>
{{{4x^2+y^2 = 4}}}
{{{x^2/1+y^2/4 = 1}}}<br>
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.<br>
Here is a graph showing the ellipse as two functions, {{{y = sqrt(4-4x^2)}}} and {{{y = -sqrt(4-4x^2))}}}<br>
{{{graph(400,400,-2,2,-3,3,sqrt(4-4x^2),-sqrt(4-4x^2))}}}<br>
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.<br>
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.<br>
Let D be the square of the distance between (1,0) and a point on the ellipse in the second quadrant.<br>
{{{D = (x-1)^2 + (4-4x^2) = -3x^2-2x+5}}}
{{{dD/dx = -6x-2}}}<br>
Find where the derivative is zero:<br>
{{{-6x-2 = 0}}}
{{{x = -1/3}}}<br>
{{{y^2 = 4-4x^2 = 4-4/9 = 32/9}}}
{{{y = (4/3)*sqrt(2)}}}<br>
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)).