Q95 / 160 · AP EAMCET 2024 (21 May Shift 2) · Mathematics
The descending order of magnitude of the eccentricities of the following hyperbolas is
A. A hyperbola whose distance between foci is three times the distance between its directrices.
B. Hyperbola in which the transverse axis is twice the conjugate axis.
C. Hyperbola with asymptotes
$x+y+1=0, x-y+3=0$
- AC, A, B
- BB, C, A
- CC, B, A
- Correct answerDA, C, B
Solution
A: Distance between foci is three times the distance between its directrices.
$\therefore 2 a e=3 \times \frac{2 a}{e} \Rightarrow e^2=3 \Rightarrow e=\sqrt{3}=1.732$
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 $A, C, B$.
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 $A, C, B$.
Hence, the correct answer is option: 'D'.