circle belonging to a ellipse

**rmpatel5** · September 23rd 2008, 02:20 PM

See attachment, thanks for looking

**icemanfan** · September 23rd 2008, 02:52 PM

The ellipse will have foci at (-d, 0) and (d, 0). Let the sum of the distances $F_1P + F_2P = c$ . Then the equation of the ellipse is $\sqrt{y^2 + (x+d)^2} + \sqrt{y^2 + (x-d)^2} = c$ . I simplified this equation to $\frac{4}{c^2}x^2 + \frac{4}{c^2 - 4d^2}y^2 = 1$ . Now can you show that the circle $x^2 + y^2 = \frac{c^2}{4}$ has the desired property?

**rmpatel5** · September 23rd 2008, 07:19 PM

Originally Posted by icemanfan

The ellipse will have foci at (-d, 0) and (d, 0). Let the sum of the distances $F_1P + F_2P = c$ . Then the equation of the ellipse is $\sqrt{y^2 + (x+d)^2} + \sqrt{y^2 + (x-d)^2} = c$ . I simplified this equation to $\frac{4}{c^2}x^2 + \frac{4}{c^2 - 4d^2}y^2 = 1$ . Now can you show that the circle $x^2 + y^2 = \frac{c^2}{4}$ has the desired property?

How did u simplify your the equation to the following $\frac{4}{c^2}x^2 + \frac{4}{c^2 - 4d^2}y^2 = 1$ .??? I tired to square both sides but it just leaves one big mess and i can never get down to the same equation you got.

**icemanfan** · September 23rd 2008, 07:34 PM

Ok, here goes:

$\sqrt{y^2 + (x+d)^2} + \sqrt{y^2 + (x-d)^2} = c$

$\sqrt{y^2 + (x+d)^2} = c - \sqrt{y^2 + (x-d)^2}$

$y^2 + (x+d)^2 = c^2 - 2c\sqrt{y^2 + (x-d)^2} + y^2 + (x-d)^2$

$(x+d)^2 - (x-d)^2 - c^2 = -2c\sqrt{y^2 + (x-d)^2}$

$x^2 + 2xd + d^2 - (x^2 - 2xd + d^2) - c^2 = -2c\sqrt<br /> {y^2 + (x-d)^2}$

$x^2 + 2xd + d^2 - x^2 + 2xd - d^2 - c^2 = -2c\sqrt{y^2 + (x-d)^2}$

$4xd - c^2 = -2c\sqrt{y^2 + (x-d)^2}$

$(4xd - c^2)^2 = 4c^2(y^2 + (x-d)^2)$

$16x^2d^2 - 8xdc^2 + c^4 = 4c^2y^2 + 4c^2(x-d)^2$

$16x^2d^2 - 8xdc^2 + c^4 = 4c^2y^2 + 4c^2(x^2 - 2xd + d^2)$

$16x^2d^2 - 8xdc^2 + c^4 = 4c^2y^2 + 4c^2x^2 - 8xdc^2 + 4c^2d^2$

$16x^2d^2 + c^4 = 4c^2y^2 + 4c^2x^2 + 4c^2d^2$

$(16d^2 - 4c^2)x^2 - 4c^2y^2 = 4c^2d^2 - c^4$

$4(4d^2 - c^2)x^2 - 4c^2y^2 = c^2(4d^2 - c^2)$

$\frac{4x^2}{c^2} - \frac{4y^2}{4d^2 - c^2} = 1$

$\frac{4x^2}{c^2} + \frac{4y^2}{c^2 - 4d^2} = 1$

**rmpatel5** · September 23rd 2008, 09:20 PM

i know i need to confirm that substituting

into the ellipse and

into the circle will result in the same values of x

this k is equal to

. How does one know that k is equal to this??

**rmpatel5** · September 24th 2008, 05:44 AM

Also, have another question. How does one know that the center of the circle in this problem is at c^2/4??

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