# Find eccentricity when asymptotes are given What is the Eccentricity of hyperbola whose asymptotes are x+2y=3 and 2x−y=1. I know for any hyperbola ((x−h)^2)/(a^2)−((y−k)^2)/(b^2)=1 The asymptotes are y−k=+- b/a (x−h) But how to solve this problem

Find eccentricity when asymptotes are given
What is the Eccentricity of hyperbola whose asymptotes are $x+2y=3$ and $2x-y=1$.
I know for any hyperbola $\frac{\left(x-h{\right)}^{2}}{{a}^{2}}-\frac{\left(y-k{\right)}^{2}}{{b}^{2}}=1$
The asymptotes are $y-k=±\frac{b}{a}\left(x-h\right)$
But how to solve this problem
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Observe that the product of the slopes of the asymptotes $-\frac{1}{2}\cdot 2=-1$ (equivalently, the dot product of the normals $⟨1,2⟩\cdot ⟨2,-1⟩=0$), which means they are orthogonal and so the hyperbola is rectangular. This is fortunate, because otherwise the problem is ambiguous. The eccentricity is therefore $\sqrt{1+{\left(\frac{1}{1}\right)}^{2}}=\sqrt{2}$.
More generally, for a hyperbola in standard position the slopes of the asymptotes are, as you’ve written, $\mathrm{tan}\theta =±\frac{b}{a}$. From this we have
$e=\sqrt{1+\frac{{b}^{2}}{{a}^{2}}}=\sqrt{1+{\mathrm{tan}}^{2}\theta }=\mathrm{sec}\theta .$
The angle between the asymptotes is $2\theta$. This gives us a couple of ways to recover the eccentricity directly from the asymptotes. One is to plug the two slopes into the formula for the tangent of the difference of two angles to get $\mathrm{tan}2\theta$ and then use the formula for the tangent of a half-angle.
Alternatively, $\mathrm{cos}2\theta$ can be recovered from the dot product of the normals or direction vectors of the asymptotes, and an application of the half-angle cosine formula gets us $e=\sqrt{\frac{2}{1+\mathrm{cos}2\theta }}$.
A third way, that might be more convenient in some situations, is to compute the angle bisector of the asymptotes and rotate so that it becomes parallel to the -axis. The resulting asymptote slopes are . There are some other ways to solve the general problem as well, but they’re even more work than this.