<s>[INST] Computing the Semimajor and Semiminor axis of an Ellipse I have the equation of the ellipse which is $\frac {x^2}{4r^2}+\frac{y^2}{r^2}=1$  Putting the (4,2) point on the ellipse we get $r^2=8$ so we get the equation $\frac {x^2}{32}+\frac {y^2}8=1$ and the semi-major axis is $\sqrt {32}= 4\sqrt 2$, the semi-minor axis is $\sqrt 8=2\sqrt 2$.
My question is using the value of  $r^2=8$  to $\frac {x^2}{4r^2}+\frac{y^2}{r^2}=1$  , how come the semi-major axis is $\sqrt {32}= 4\sqrt 2$ ? and semi-minor axis is $\sqrt 8=2\sqrt 2$? 
I am puzzled on how to get that answer.
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anyway this question is related to my previous question Finding the Width and Height of Ellipse given an a point and angle
 [/INST] A circle centred at the origin, of radius $r$ has equation $x^2 + y^2 = r^2$ (alternatively, $\frac{x^2}{r^2} + \frac{y^2}{r^2} = 1$).
An ellipse centred at the origin has equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where $a, b > 0$. The larger of the two numbers $a$ and $b$ is called the (length of) the semi-major axis, while the smaller of the two numbers is called the (length of) the semi-minor axis.
Note, the case where the (lengths of the) semi-major and semi-minor axes are the same is precisely when the ellipse is a circle.
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