Computing the Semimajor and Semiminor axis of an Ellipse

Question

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.

... anyway this question is related to my previous question Finding the Width and Height of Ellipse given an a point and angle

Answer 1

The ellipse is in standard position and orientation, with the axes along the $x$ and $y$ axes. The ellipse meets the $x$-axis at the two points $x=\pm\sqrt{32}$, $y=0$, and meets the $y$-axis at $x=0$, $y=\pm\sqrt{8}$.

The distance from $(-\sqrt{32},0)$ to $(\sqrt{32},0)$ is bigger than the distance between the two $y$-intercepts.

The major axis therefore has length $2\sqrt{32}$, and the semi (half) major axis is half of that.

Answer 2

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.