Q108 / 180 · KCET 2013 · Mathematics
If the distance between the focii and the distance between the directrices of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ are in the ratio $3: 2$, then $a: b$ is
- Correct answerA$\sqrt{2}: 1$
- B$1: 2$
- C$\sqrt{3}: \sqrt{2}$
- D$2: 1$
Solution
Given equation of hyperbola is $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$
Now, distance between the focii $=2 a e$ and distance between the directrices $=\frac{2 a}{e}$ (by condition)
Given, ratio $=3: 2$
$\frac{2 a e}{(2 a / e)}=\frac{3}{2}$
$\Rightarrow \quad 4 a e^{2}=6 a$
$\Rightarrow \quad e^{2}=\frac{3}{2} \Rightarrow e=\sqrt{\frac{3}{2}}...(i)$
$\because$ Eccentricity of hyperbola $=\sqrt{\frac{a^{2}+b^{2}}{a^{2}}}$
$\therefore \quad \sqrt{\frac{a^{2}+b^{2}}{a^{2}}}=\sqrt{\frac{3}{2}} \quad$ [from Eq. (i)]
$\Rightarrow \quad \frac{a^{2}+b^{2}}{a^{2}}=\frac{3}{2}$ (squaring on both sides)
$\Rightarrow \quad 1+\frac{b^{2}}{a^{2}}=\frac{3}{2}$
$\Rightarrow \quad\left(\frac{b}{a}\right)^{2}=\frac{1}{2}$
$\therefore \quad \frac{b}{a}=\frac{1}{\sqrt{2}}$
Hence, $\quad a: b=\sqrt{2}: 1$
Hence, the correct answer is option: 'A'.