# Hyperbola with Perpendicular Asymptotes

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?

• Just above your bold faced "Case 1", you have the symbol for parallel instead of perpendicular. Otherwise your reasoning is just fine. – B. Goddard Mar 12 '18 at 11:18
• Oh yes, that's right. Thanks for the help! – AppleCrazy Mar 13 '18 at 3:10
• 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: $\sqrt{2}$. – Blue Mar 13 '18 at 3:20
• Wow, that's much faster, now that I think about it. – AppleCrazy Mar 13 '18 at 4:48