I was working through conics problems, and I came across an interesting question:

How do the values of a and b in the standard form of the equation of a hyperbola relate if the asymptotes of a hyperbola are perpendicular? What is the eccentricity of these type of hyperbolas?

I immediately looked for a relation between a and b using the slopes of the asymptotes:

$$ m_1m_2=-1 \text{, if perpendicular lines} $$ $$ \textbf{Case I: Hyperbolas are vertical}$$ $$ (\frac{a}{b})(-\frac{a}{b}) = -1 $$ $$ \frac{-a^2}{b^2} = -1$$ $$ \therefore a^2 = b^2 $$ $$ \textbf{Case II: Hyperbolas are horizontal}$$ $$ (\frac{b}{a})(-\frac{b}{a}) = -1 $$ $$ \frac{-b^2}{a^2} = -1$$ $$ \therefore a^2 = b^2 $$

I claimed on the basis of these results that the squares of a and b are equal, but is that a valid point to make and/or is there another relation that I'm missing?

For the second part of the question, I knew that the eccentricity of a hyperbola is given by $ e = c \div a $ and that a, b, and c are related by $ c^2 = a^2 + b^2 $. Based on these equations and the equality established in the previous step:

$$ e = \frac{c}{a} = \frac{\sqrt{a^2 + b^2}}{a} = \frac{\sqrt{a^2 + a^2}}{a} = \frac{\sqrt{2a^2}}{a} = \frac{\sqrt{2}a}{a} = \sqrt{2}$$

Thus, the eccentricity is $\sqrt{2}$. But this answer is based on the assumption that my first step is correct, so is this proof valid for all hyperbolas such that their asymptotes are perpendicular?

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    $\begingroup$ Just above your bold faced "Case 1", you have the symbol for parallel instead of perpendicular. Otherwise your reasoning is just fine. $\endgroup$ – B. Goddard Mar 12 '18 at 11:18
  • $\begingroup$ Oh yes, that's right. Thanks for the help! $\endgroup$ – AppleCrazy Mar 13 '18 at 3:10
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    $\begingroup$ Without having to reference a particular orientation, you could simply note that the asymptotes of a hyperbola with transverse axis $2a$ and conjugate axis $2b$ are diagonals of an $2a$-by-$2b$ rectangle. If those diagonals are perpendicular, then that rectangle is a square, so that $a=b$. Eccentricity, then, is as you've computed: $\sqrt{2}$. $\endgroup$ – Blue Mar 13 '18 at 3:20
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    $\begingroup$ Wow, that's much faster, now that I think about it. $\endgroup$ – AppleCrazy Mar 13 '18 at 4:48

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