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Find the points on the ellipse 4x2 + y2 = 4 that are farthest away from the point (-1, 0).
Solution:
Consider (x, y) as the point on the ellipse 4x2 + y2 = 4
We can write it as
y2 = 4 - 4x2 --- (1)
\(y=\pm 2\sqrt{1-x^{2}}\)
Distance between (x, y) and (1, 0) is
\(d(x)=\sqrt{(x-1)^{2}+y^{2}}\)
Substituting equation (1)
\(d(x)=\sqrt{(x-1)^{2}+4-4x^{2}}\)
By further calculation
\(d(x)=\sqrt{-3x^{2}-2x+5}\)
Maximize f (x) = -3x2 - 2x + 5
f’(x) = - 6x - 2 = 0
The only critical value is x = - 1/3
f’’(x) = - 6
Where x = -1/3 which maximizes f(x) and d(x)
\(y=\pm 2\sqrt{1-(\frac{-1}{3})^{2}}=\pm \frac{4\sqrt{2}}{3}\)
Therefore, the farthest points away from the point (-1, 0) are (-1/3, ±4√2/3).
Find the points on the ellipse 4x2 + y2 = 4 that are farthest away from the point (-1, 0).
Summary:
The points on the ellipse 4x2 + y2 = 4 that are farthest away from the point (-1, 0) are (-1/3, ±4√2/3).
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