When should I shift $a$ and $b$ in $\cfrac{x^2}{a^2}+\cfrac{y^2}{b^2}=1$?

Question

Find the reduced equation of the elypsis such that: The foci are $(0,6);(0,-6)$ and the larger axis has length $34$.

I did the following:

Taking the equation $\cfrac{x^2}{a^2}+\cfrac{y^2}{b^2}=1$, with $b^2=a^2-c^2$.

With the information given above, then the focal distance is $2c=12$ then $c=6.$ As the larger axis has length $34$, $a=17$, then $b^2=289-36=253$ and then $b=\sqrt{253}$.

The equation of the conic should be:

$$\cfrac{x^2}{289}+\cfrac{y^2}{253}=1$$

But it is:

$$\cfrac{x^2}{253}+\cfrac{y^2}{289}=1$$

According to the book answers. I understand that this conic has the numbers $a$ and $b$ shifted, I guess that it's because the larger axis is on the $y$-axis but I'm not sure if this is the rule to shift $a$ and $b$ because some time ago I did one exercise and this theory was invalid (assuming I got it correct). So, when shoul I actually shift $a$ and $b$?

You are right, this is because the larger axis is on the y-axis. — Ofir Schnabel, Jul 01 '15 at 08:05
@OfirSchnabel Oh. Then I probably made some mistake in the previous computations. Gonna try it now. — Red Banana, Jul 01 '15 at 08:08
The foci are always on the larger axis and the $a$ is always "with respect" to the larger axis. This is also true for the non-canonical case. — Ofir Schnabel, Jul 01 '15 at 08:10

score 1 · Accepted Answer · answered Jul 01 '15 at 08:48

1

The foci are (0,6);(0,-6). Hense, they are located on the y-asis. So, the larger axis is on y-axis and $b=17$.

As a consequence, $c^2=b^2-a^2$, but not $c^2=a^2-b^2$ as you did.

With $a^2=b^2-c^2$ you will obtain the expected result.

answered Jul 01 '15 at 08:48

JJacquelin

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When should I shift $a$ and $b$ in $\cfrac{x^2}{a^2}+\cfrac{y^2}{b^2}=1$?

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