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Find the equations of the following conics, each with its centre at the origin.

(a) A hyperbola with foci $(\pm4, 0)$ and directrices $x= \pm2$

(b) An ellipse with foci $(0, \pm4)$ and directrices $y= \pm6$.

Probably some very very simple solutions but it has left me scratching my head as to how to find the equations...?

Gerry Myerson
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amjh23
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  • What have you tried? How do you think your instructor expects you to solve this problem? Have you at least tried sketching the figures? – David H Oct 09 '14 at 23:22
  • I have looked at some info he has provided the class such as starting with the equation and from there finding the vertex, foci, eccentricity and directrices. However I can't seem to get the equation when working backwards. It is expected we find the solution through algebra – amjh23 Oct 10 '14 at 00:39

1 Answers1

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According to Wikipedia, any point on a conic has the same ratio of distances to focus and directrix. You don't know that ratio, but since you know two foci and directrices, you can simply set them equal. To avoid square roots, it's better to take the square of that, i.e. the ratio of squared distances.

So for (a) you want

$$\frac{(x-4)^2+y^2}{(x-2)^2} = \frac{(x+4)^2+y^2}{(x+2)^2}$$

and for (b) you want

$$\frac{x^2+(y-4)^2}{(y-6)^2} = \frac{x^2+(y+4)^2}{(y+6)^2}$$

Cross-multiply, simplify, factor out the line which appears in each of these results, and you have the canonical implicit notation of each conic.

MvG
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