"Find the points on the ellipse 4x2+y2=4 that..." was answered by our community

Tepliwodav0

Answered question

2021-12-01

Find the points on the ellipse

4 x^{2} + y^{2} = 4

that are the farthest from the point (1;0)

Answer & Explanation

Mespirst

Beginner2021-12-02Added 17 answers

Let (x;y) be the points on the ellipse

4 x^{2} + y^{2} = 4

Then,

4 x^{2} + y^{2} = 4 r i g h t \Rightarrow y^{2} = 4 - 4 x^{2} r i g h t \Rightarrow y = \pm 2 \sqrt{1 - x^{2}}

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

If

y^{2} = 4 - 4 x^{2}

, then we have:

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

\sqrt{- 3 x^{2} - 2 x + 5}

Then maximise

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

f^{'} (x) = - 6 x - 2 = 0 \Rightarrow - \frac{1}{3}

(the only critical value)

f^{″} (x) = - 6 \Rightarrow x = - \frac{1}{3}

Since

y = \pm 2 \sqrt{1 - {(- \frac{1}{3})}^{2}} = \pm \frac{4 \sqrt{2}}{3}

The farthest points are

(- \frac{1}{3}; \pm \frac{4 \sqrt{2}}{3})

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