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# Find the points on the ellipse 4x^{2} + y^{2} = 4 that are farthest away from the point (-1, 0).

**Solution:**

Consider (x, y) as the point on the ellipse 4x^{2} + y^{2} = 4

We can write it as

y^{2} = 4 - 4x^{2} --- (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) = -3x^{2} - 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 4x^{2} + y^{2} = 4 that are farthest away from the point (-1, 0).

**Summary:**

The points on the ellipse 4x^{2} + y^{2} = 4 that are farthest away from the point (-1, 0) are (-1/3, ±4√2/3).

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