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I have worked on this one for a while and I can not make my answer match the author's.

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

I have:

$$4x^2 + y^2 = 4$$

and then the distance formula, so I set y to terms of x and I get

$$\sqrt{(x-1)^2 + (2-2x)^2}$$

Setting the difference the a square this gives me a derivative of

$$10x-10$$

which gives me a zero of 1, this is wrong according to the book and I am not sure why.

  • 2
    Of course it gives you that the minimum is when $x = 1$. You've just said that the point closest to $(1,0)$ is, in fact, $(1,0)$! But you haven't used that the points need to be on an ellipse.2012-04-04
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    Did you do $\sqrt{4-4x^2}=\sqrt4-\sqrt{4x^2}$? Please tell me you didn't do that....2012-04-04

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