What Curve Results from the Centers of Perpendicular Lines to an Ellipse?

[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

please help

 

[HallsofIvy]

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

 

[transgalactic]

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

 

[transgalactic]

how can i solve this thing?

 

[HallsofIvy]

$x^2+ 2y^2= 8$ 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:
$\frac{x^2}{8}+ y^2=1$.

 

[HallsofIvy]

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

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=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)

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

in my book it gives a different answer

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

please help

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]

 

[transgalactic]

thank you very much

 

[HallsofIvy]

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