find points on ellipse 4x^2 + y^2 = 4 furthest from (1, 0)

Limits, differentiation, related rates, integration, trig integrals, etc.
nona.m.nona
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find points on ellipse 4x^2 + y^2 = 4 furthest from (1, 0)

Find the points on the ellipse 4x2 + y2 = 4 that are farthest away from the point (1, 0).

I must be forgetting something easy, to get started.

stapel_eliz
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I must be forgetting something easy...
The Distance Formula, maybe (or, perhaps more usefully, the square of it), plus a little solving...?

Don't forget that the points on the ellipse can be stated in terms of "y=", but you have to use two "halves":

. . . . .$4x^2\, +\, y^2\, =\, 4$

. . . . .$y^2\, =\, 4\, -\, 4x^2$

. . . . .$y\, =\, \pm 2 \sqrt{1\, -\, x^2}$

Then take a "half" (say, the "minus" half), note that the points on the ellipse are of the form:

. . . . .$(x,\, y)\, =\, (x,\, -2 \sqrt{1\, -\, x^2}$

...and then plug the two "points" into the (square of the) Distance Formula:

. . . . .$d(x)\, =\, \left(x\, -\, 1\right)^2\, +\, \left(-2 \sqrt{1\, -\, x^2}\, -\, 0\right)^2$

. . . . . . . . ..$=\, x^2\, -\, 2x\, +\, 1\, +\, 4(1\, -\, x^2)$

. . . . . . . . ..$=\, x^2\, -\, 4x^2\, -\, 2x\, +\, 1\, +\, 4$

...and so forth.

Eliz.

nona.m.nona
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Re: find points on ellipse 4x^2 + y^2 = 4 furthest from (1, 0)

So I get a "distance" function of $d(x)\, =\, 5\, -\, 2x\, -\, 3x^2$. The max/min points are the critical points, so $d'(x)\, =\, -2\, -\, 6x\, =\, 0$, or x = -1/3.

For x = 3, I get $4\left(-\frac{1}{3}\right)^2\, +\, y^2\, = 4$, so $y\, =\, \pm \frac{4\sqrt{2}}{3}$.

I think the other "half" gives the same answer, because of the squaring. Does this look right?

stapel_eliz
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Does this look right?
Checking the graph (using decimal approximations for y, of course), your solutions look sensible.

Eliz.

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