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For an ellipse, $9x^2+4y^2=36$, find vertices and foci.

I would first standardised the equation in form $(x)^2/a^2 + y^2/b^2=1$ thus…………(i)

divide all sides by 36,I get: $$(9x^2)/36+(4y^2)/36=36/36$$ which is equiv. to $x^2/4+y^2/9=36/36$

based on $x^2/4+y^2/9=36/36$ we can rewrite this as: $$x^2/2^2 +y^2/3^2 =1$$ which is in standard $x^2/a^2 +y^2/b^2 =1$

we can even pluck out our values for a,b thus: a=2 and b=3

but by pythagoras,we know $c^2=a^2-b^2$ hence $c^2=2^2-3^2$

so in my case, $c^2=4-9= -5$ and $\therefore$ $c=√(-5)$

But I reckon the values of a,b can swap depending on major axis since we can re-write standard equation as $$(x)^2/b^2 + y^2/b^2=1$$ which will bring a different value of $c^2$.

Questions: How do I make sure I get the correct value of $c^2$? Question: How do ensure the ellipse is plotted on the correct major axis?

Sylvester
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  • If I swap the values I would rewrite $x^2/b^2 + y^2/a^2=1$. Thus $x^2/3^2+y^2/2^2 = 1$. This will give $$c^2 = 3^2-2^2 = \sqrt(5)$$ – Sylvester Sep 17 '13 at 04:58
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    First, 36/36 is just 1. – The Chaz 2.0 Sep 17 '13 at 04:59
  • Correct. Sorry let me correct that in my answer. – Sylvester Sep 17 '13 at 05:02
  • To make sure that you found correct value of $c$, or any other "geometric" value, just check if it is real, not complex like $\sqrt {-5}$. – Kaster Sep 17 '13 at 05:05
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    ^That's not correct, @Kaster. For an ellipse, $a$ is the larger of a, b, not the one that goes with the $x$. – The Chaz 2.0 Sep 17 '13 at 05:16
  • @TheChaz2.0 You can think of it that way too. For me algebraic representation of an ellipse as $x = a \cos \omega t, y = b \sin \omega t$ comes first, so $a$ is "whatever goes with the $x$", not a conic section. – Kaster Sep 17 '13 at 05:22

1 Answers1

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For the ellipse $9x^2 + 4y^2 = 36$ ...

This is, as you have found, equivalent to $$\dfrac{x^2}{4} + \dfrac{y^2}{9} = 1$$

Now in an ellipse, the larger denominator corresponds with $a$.
(Contrast this with a hyperbola, where the denominator of the positive variable term is $a^2$)

So we find that $a = 3, b = 2$. The major axis is vertical, as the $a$ was under the $y^2$.

Therefore, the vertices are up and down $(a = 3)$ units from the center, which is the origin in this case.
The covertices are $(b = 2)$ units left and right of center.

$c = \sqrt5$, and the foci are $c$ units up and down from the center.

The Chaz 2.0
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    in an ellipse, the larger denominator corresponds with a. Puzzling. I would put it this way: in the standard form $x^2/a^2 + y^2/b^2 = 1$, it doesn't matter whether $a$ or $b$ is larger. You get an ellipse either way. The $a$ goes with the $x$ in the sense that the ellipse crosses the $x$-axis where $x = \pm a$. It's the formulae for foci and eccentricity that need some adjustment.

     – bubba Sep 17 '13 at 05:34
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    @bubba - hope it's not too puzzling, as the precalculus books that I've used for 10 years (Larson & Hostetler, pg 744) have exactly what I have written. – The Chaz 2.0 Sep 17 '13 at 05:40
  • @TheChaz2.0, I want to get it clear:If I get an equation $$x^2/2^2 +y^2/3^2 = 1$$: To know which value is $a$, I check the denominator is greater which in this case is $3^2$. To know which is the major axis, I check under which $a$ is the denominator, which is on $y^2$. The axis will be $x$. How do I know whether it is vertical or horizontal? – Sylvester Sep 17 '13 at 08:19