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by **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:

**Sponsor**

by **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":

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

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

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

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.

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

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?

by **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:

Eliz.

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

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

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