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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.

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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

1 Answers 1

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We want to maximize $\sqrt{(x-1)^2+(y-0)^2}$, given that $4x^2+y^2=4$.

Equivalently, we want to maximize $(x-1)^2+y^2$, same side condition.

But $y^2=4-4x^2$.

So we want to maximize $(x-1)^2+(4-4x^2)$. Your turn.

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    I thought $y^2=2-2x^2$ so I just took the square root of that which I do not see as being wrong because I can't take the square of a negative.2012-04-04