Stretched Ellipse: what is its equation?

D!ddy

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Nov 2, 2007
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19
I need help with this question please,
The circle x[sup:1jnbnj5r]2[/sup:1jnbnj5r] + y[sup:1jnbnj5r]2[/sup:1jnbnj5r] - 2x - 3 = 0 is stretched horizontally by a factor of 2 about x = 0 to obtain an ellipse. What is the equation of this ellipse in general form?

I converted it to standard form and got
(x-1)[sup:1jnbnj5r]2[/sup:1jnbnj5r]/4 + y[sup:1jnbnj5r]2[/sup:1jnbnj5r]/4 = 1

If i wanted to stretch it horizontally by a factor of 2, do i multiply the 4 under (x-1)[sup:1jnbnj5r]2[/sup:1jnbnj5r] by 2 then convert the whole equation back to general form?
 
D!ddy said:
I need help with this question please,
The circle x[sup:2c9p1f2d]2[/sup:2c9p1f2d] + y[sup:2c9p1f2d]2[/sup:2c9p1f2d] - 2x - 3 = 0 is stretched horizontally by a factor of 2 about x = 0 to obtain an ellipse. What is the equation of this ellipse in general form?
you sure it isn't stretched about y = 0 instead?
 
Hello, D!ddy!

\(\displaystyle \text{The circle }x^2 + y^2 - 2x - 3 \:= \:0\text{ is stretched horizontally by a factor of 2 \;{\bf about }x = 0}}\)

\(\displaystyle \text{to obtain an ellipse. What is the equation of this ellipse in general form?}\)

If no one is responding, it's probably because we're not sure what "about x = 0" means.

\(\displaystyle \text{The circle is: }\:(x-1)^2 + y^2\:=\:4\text{, with center (1,0) and radius 2.}\)


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\(\displaystyle \text{If I were to }guess\text{ what they meant, I would say:}\)

. . \(\displaystyle \text{the left }x\text{-intercept }(-1,0)\text{ is moved to }(-2,0)\)
. . \(\displaystyle \text{the center is moved from }(1,0)\text{ to }(2,0)\)
. . \(\displaystyle \text{the right }x\text{-intercept }(3,0)\text{ is moved to }(6,0)\)

\(\displaystyle \text{This is an ellipse with center at }(2,0)\text{ and: }a = 4,\;b = 2\)

\(\displaystyle \text{Its equation is: }\;\frac{(x-2)^2}{16} + \frac{y^2}{4} \;=\;1\)

 
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