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

Limits, differentiation, related rates, integration, trig integrals, etc.

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

Postby nona.m.nona on Fri Feb 06, 2009 7:54 pm

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. :roll:
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Postby stapel_eliz on Fri Feb 06, 2009 9:11 pm

nona.m.nona wrote:I must be forgetting something easy...

The Distance Formula, maybe (or, perhaps more usefully, the square of it), plus a little solving...? :wink:

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

. . . . .

. . . . .

. . . . .

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

. . . . .

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

. . . . .

. . . . . . . . ..

. . . . . . . . ..

...and so forth. :D

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

Postby nona.m.nona on Mon Feb 09, 2009 2:41 pm

So I get a "distance" function of . The max/min points are the critical points, so , or x = -1/3.

For x = 3, I get , so .

I think the other "half" gives the same answer, because of the squaring. Does this look right?
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Postby stapel_eliz on Mon Feb 09, 2009 5:14 pm

nona.m.nona wrote:Does this look right?

Checking the graph (using decimal approximations for y, of course), your solutions look sensible. :thumb:

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