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

**Solution:**

Given, equation of ellipse is 4x^{2} + y^{2} = 4

We have to find the points on the ellipse that are farthest away from the point (1, 0)

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)

Taking square root,

y = \(\pm 2\sqrt{1-x^{2}}\)

Distance between (x, y) and (1, 0) is

d(x) = \(\sqrt{(x-1)^{2}+y^{^{2}}}\)

Substituting the value of y²,

d(x) = \(\sqrt{(x-1)^{2}+4-4x^{2}}\)

On simplification,

d(x) = \(\sqrt{x^{2}+1-2x+4-4x^{2}}\)

d(x) = \(\sqrt{-3x^{2}-2x+5}\)

Now we have to maximize f(x)

f(x) = \(\sqrt{-3x^{2}-2x+5}\)

f’(x) = -6x - 2

To find critical value,f’(x) = 0

-6x - 2 = 0

-6x = 2

x = -2/6

x = -1/3

Now, f’’(x) = -6

Where x=-1/3 which maximizes f(x) and d(x)

\(y=\pm 2\sqrt{1-(\frac{-1}{3})^{2}}\)

\(y=\pm 2\sqrt{1-(1/9)}\\y=\pm 2\sqrt{\frac{8}{9}}\)

\(y=\pm \frac{2(2\sqrt{2})}{3}\)

\(y=\pm \frac{4\sqrt{2}}{3}\)

Therefore, the farthest points away from the point (1, 0) are (-1/3 , \(y=\pm \frac{4\sqrt{2}}{3}\))

## Find the points on the ellipse 4x^{2} + y^{2} = 4 that are farthest away from 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, \(y=\pm \frac{4\sqrt{2}}{3}\))

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