The descending order of magnitude of the eccentricities of the following hyperbolas is A. A hyperbola whose…
- C, A, B
- B, C, A
- C, B, A
- A, C, B
Solution
B : The transverse axis is twice the conjugate axis $\therefore 2 a=2(2 b) \Rightarrow a=2 b \Rightarrow b=\frac{a}{2}$
We know that $a^2 e^2=a^2+b^2=a^2+\frac{a^2}{4}$ $e^2=\frac{5}{4} \Rightarrow e=\frac{\sqrt{5}}{2}=1.11$
C : Slope of asymptotes are $m_1=-1$ and $m_2=1$
$\begin{aligned} & \therefore m_1 \cdot m_2=-1, \text { so it is rectangular hyperbola } \\ & \Rightarrow e=\sqrt{2}=1.414 \end{aligned}$
Hence order is $A, C, B$.
Asked in: AP EAMCET 2024 (21 May Shift 2)