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One focus and the nearest directrix to this focus are $(0,-5)$ and $d_1: x-3y-4=0$, respectively. Also, the eccentricity is $\sqrt{10}$. Find the second directrix of the hyperbola.

I assumed the second directrix is $d_2: x-3y+c=0$ as directrices are parallel. And then, I got two possible values for $c$, namely, $-\cfrac{14}{9}; -\cfrac{256}{9}$.

How to decide which to choose?

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One of your roots corresponds to a hyperbola, which is what you want. The second root corresponds to an ellipse. The ellipse in the second root has an eccentricity equal to $1$ divided by the eccentricity of the hyperbola you want. A key difference between the two conic sections is that a hyperbola has its directrices between the foci, while for an ellipse it's the other way around.

To orient yourself, find the value of $c$ such that $x-3y+c=0$ at the given focus. This, of course, is $-15$. The given directrix has $c=-4$, which is greater. So now, if you try the root $c=-256/9<-15$ for the other directrix, you have the focus between the directrices which corresponds to that ellipse. To get the hyperbola you must select $c>-4$ for the other directrix, putting the given directrix ($c=-4$) between the focus ($c=-15$) and the directrix you're trying to find. Therefore select $c=-14/9$.

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