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
Best Answer
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.