An ellipse question?

1. Feb 19, 2007

transgalactic

i added a file with the curve the is being asked to find

an ellipse is given. its formula is x^2 + 2*y^2=8 .

find the formula of the curve that is being created by the

the centers of the perpendicular lines to the X axes and the ellipse.

i tried to solve this question
by putting the Y^2 on the one side and on the other the rest
and devidind it by 2 .

2y^2=8-x^2
y^2=8-x^2/2
y=V(8-x^2/2)
(v=root simbol)

the curve that in the center always smaller in height by 2
so i devided the formula by 2 to find our curve

y=1/2*V(8-x^2/2)
in my book it gives a different answer

(x^2)/8 +y^2=1

plz help

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2. Feb 19, 2007

HallsofIvy

Staff Emeritus
What do you mean by "the centers of the perpendicular lines to the X axes and the ellipse"?

3. Feb 19, 2007

transgalactic

like it shows at the pictue
a curve wich composed from the middle points of this straight lines
parrallel to the Y axes

4. Feb 20, 2007

transgalactic

how can i solve this thing???

5. Feb 20, 2007

HallsofIvy

Staff Emeritus
$x^2+ 2y^2= 8$ is the same as $\frac{x^2}{8}+ \frac{y^2}{4}= 1$.
That's an ellipse with major semi-axis, along the x-axis, of length $\sqrt{8}= 2\sqrt{2}$ and minor semi-axis, along the y-axis, of length 2.

Dividing the y-coordinate of each point by 2 gives an ellipse with the same major semi-axis but minor semi-axis of length 1:
$\frac{x^2}{8}+ y^2=1$.

6. Feb 20, 2007

HallsofIvy

Staff Emeritus
Here's your error! Don't know why I didn't see this sooner. Dividing both sides of 2y2= 8- x2 by 2 gives y2= 4- x2!

$y= 1/2 \sqrt{4- x^2}$

Square both sides of $2y= \sqrt{4- x^2}$ and you get
$4y^2= 4- x^2$ or $x^2+ 4y^2= 4$.

Divide through by 4:
[tex]\frac{x^2}{4}+ y^2= 1[/itex]

7. Feb 21, 2007

transgalactic

thank you very much

8. Feb 22, 2007

HallsofIvy

Staff Emeritus
Now, my question is, "what does this have to do with 'Tensor Analysis and Differential Geometry'?"