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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

Community
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Chinchan Zu
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2 Answers2

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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.

André Nicolas
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  • hi andre, thanks for the explanation. I mean a user gave me an answer with like a squareroot of 32... but what i dont understand is how did he computed it using the equation. Im not really good at math so i dont know what are the steps for substitution. thanks man – Chinchan Zu Apr 22 '13 at 08:24
  • @ChinchanZu: To see where we meet the $x$-axis, put $y=0$ in the equation. Then from the equation of the ellipse we get $\frac{x^2}{32}=1$, which is equivalent to $x^2=32$. Take square roots. We get $x=\sqrt{32}$. This can be left as is, or approximated with a calculator. We can also simplify algebraically, since $\sqrt{32}=\sqrt{(4^2)(2)}=4\sqrt{2}$. – André Nicolas Apr 22 '13 at 08:37
  • thanks a bunch man! ill try this – Chinchan Zu Apr 22 '13 at 09:03
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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.

Michael Albanese
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  • hi michael! thanks for your response. A member gave me an answer which is the squareroot of 32... but i dont understand howd he computed it using the formula. can you be kind enough to show me how he lead to the answer? thanks a lot man – Chinchan Zu Apr 22 '13 at 08:19
  • For the ellipse you are dealing with, $a^2 = 4r^2$ and $b^2 = r^2$. Now, as $r^2 = 8$ is given, we have $a^2 = 4\times 8 = 32$ and $b^2 = 8$. Taking square roots, and remembering that $a, b > 0$, we have $a = \sqrt{32}$ and $b = \sqrt{8}$. – Michael Albanese Apr 22 '13 at 08:30