# Thread: Ellipse! help!

1. ## Ellipse! help!

Hi there!
um.. I am having trouble with a question I started some work and then I am not sure if I am doing it right! and if I am I don't know what to do next!

anyways here is the question that stumps me!
An ellipse is formed by strecthing the grapgh of x^2+ y^2=1 horizontally by a factor of 3 and vertically by a factor of 4. Determine the equation of this ellipse in standard form.

so my work:
x^2+ y^2=1
(x/3)^2 +(y/4)^2=1

x^2/9 + y^2/16=1

am I going on the right track or am I totally off? if I am doing this right were do I go from here?

2. Originally Posted by cutie4ever
Hi there!
um.. I am having trouble with a question I started some work and then I am not sure if I am doing it right! and if I am I don't know what to do next!

anyways here is the question that stumps me!
An ellipse is formed by strecthing the grapgh of x^2+ y^2=1 horizontally by a factor of 3 and vertically by a factor of 4. Determine the equation of this ellipse in standard form.

so my work:
x^2+ y^2=1
(x/3)^2 +(y/4)^2=1

x^2/9 + y^2/16=1

am I going on the right track or am I totally off? if I am doing this right were do I go from here?
The graph,
$\displaystyle x^2+y^2=1$
The radius is 1 of the circle.

If you stretch the x-axis by 3 you then have a semi-axis along x to be 3. Similarly for the y axis.
In that case the equation of the ellipse is,
$\displaystyle \frac{x^2}{3^2}+\frac{y^2}{4^2}=1$

3. If a is the width of the elipse, and b is the height, the elipse equation is $\displaystyle \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1$.