Hyperbola 1 Question 7
8.
Let $0<\theta<\frac{\pi}{2}$. If the eccentricity of the hyperbola $\frac{x^{2}}{\cos ^{2} \theta}-\frac{y^{2}}{\sin ^{2} \theta}=1$ is greater than 2 , then the length of its latus rectum lies in the interval
(2019 Main, 9 Jan I)
(a) $1, \frac{3}{2}$
(b) $(3, \infty)$
(c) $\frac{3}{2}, 2$
(d) $(2,3)$
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Answer:
Correct Answer: 8. (b)
Solution:
- For the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$,
$ e=\sqrt{1+\frac{b^{2}}{a^{2}}} $
$\therefore$ For the given hyperbola,
$ e=\sqrt{1+\frac{\sin ^{2} \theta}{\cos ^{2} \theta}}>2 $
$\Rightarrow 1+\tan ^{2} \theta>4$
$ \left(\because a^{2}=\cos ^{2} \theta \text { and } b^{2}=\sin ^{2} \theta\right) $
$\Rightarrow \tan ^{2} \theta>3$
$\Rightarrow \tan \theta \in(-\infty,-\sqrt{3}) \cup(\sqrt{3}, \infty)$
$ \left[x^{2}>3 \Rightarrow|x|>\sqrt{3} \Rightarrow x \in(-\infty,-\sqrt{3}) \cup(\sqrt{3}, \infty)\right] $
But $\theta \in 0, \frac{\pi}{2} \Rightarrow \tan \theta \in(\sqrt{3}, \infty)$
$\Rightarrow \theta \in \frac{\pi}{3}, \frac{\pi}{2}$
Now, length of latusrectum
$ =\frac{2 b^{2}}{a}=2 \frac{\sin ^{2} \theta}{\cos \theta}=2 \sin \theta \tan \theta $
Since, both $\sin \theta$ and $\tan \theta$ are increasing functions in $\frac{\pi}{3}, \frac{\pi}{2}$.
$\therefore$ Least value of latusrectum is
$ =2 \sin \frac{\pi}{3} \cdot \tan \frac{\pi}{3}=2 \cdot \frac{\sqrt{3}}{2} \cdot \sqrt{3}=3 \quad \text { at } \theta=\frac{\pi}{3} $
and greatest value of latusrectum is $<\infty$
Hence, latusrectum length $\in(3, \infty)$
