Circle $x^2+y^2=2$ is stretched by a scale factor $2$ parallel to the $x$-axis, find the equation of Ellipse

Question

What is the quick method or formula to finding this answer? Also the method for finding the answer when the stretch is parallel to the $y$-axis, Regards Tom

Answer 1

Let the circle have the old values of $x$ and the ellipse (stretched circle) the new values of $X$. $X=2x$, or $x=X/2$.
$x^2+y^2=1$, so $(X/2)^2+y^2=2$, so the equation of the stretched circle is $(X^2/4)+y^2=2$.
Now forget about the circle, and $X$ versus $x$, the equation is $(x^2/4)+y^2=2$.

Answer 2

If you stretch along an axis parallel to x- or y- axis, the semi-major/minor axis in that

principal direction is increased by that factor.

Stretched along $x$-axis : $\bigl(\frac{x}{2}\bigr)^2 + \bigl(\frac{y}{1}\bigr)^2 = 1$

Stretched along $y$-axis : $\bigl(\frac{x}{1}\bigr)^2 + \bigl(\frac{y}{2}\bigr)^2 = 1$

This happens to all curves with such two-fold polar symmetry... the characteristic dimension (like a,b) can be scaled up or down.